JAMB UTME - Chemistry - 1997
Ajụjụ 1 Ripọtì
The IUPAC nomenclature of the organic compound with the above structural formula is
Akọwa Nkọwa
The organic compound with the given structural formula is a branched chain alkane with a total of eight carbon atoms. To name the compound using IUPAC nomenclature, we need to identify the longest continuous carbon chain, which in this case is five carbons long. We number the carbon chain from the end that gives the substituent with the lowest possible number. In this case, we can number the chain from either end but we get the numbering 2, 5 for the substituents. The compound has one ethyl group (a two-carbon chain attached to the second carbon atom of the main chain) and two methyl groups (one attached to the second carbon atom and another to the fifth carbon atom of the main chain). Therefore, the IUPAC name of the compound is 3-ethyl-2,5-dimethylhexane. Thus, the correct option is: 3-ethyl-2,5-dimethylhexane.
Ajụjụ 2 Ripọtì
Use the graph above to answer this question.
A sample, X, solid at room temperature, was melted, heated to a temperature of 358 K and allowed to cool as shown in OPQR. The section OP suggests that X is in the
Akọwa Nkọwa
The graph shows the temperature changes of a sample X as it was melted, heated to 358 K, and allowed to cool. The section OP of the graph represents a horizontal line at a temperature of 358 K. This indicates that during the time period represented by OP, the temperature of sample X did not change. Therefore, the sample X must have been in a state where its temperature was constant, which is either the solid or liquid state. However, since the sample was melted before being heated to 358 K, it must have been in the liquid state during the time period represented by OP. Therefore, the correct answer is that X is in the liquid state during section OP.
Ajụjụ 3 Ripọtì
mE + nF → pG + qH. In the equation above, the equilibrium constant is given by?
Akọwa Nkọwa
The equation provided is a representation of a chemical reaction, where m, n, p, and q are coefficients that represent the number of moles of each substance involved in the reaction. The equilibrium constant (Kc) is the ratio of the concentrations of the products (G and H) to the concentrations of the reactants (E and F), each raised to the power of their respective coefficients. Therefore, the correct equation for the equilibrium constant (Kc) for the given chemical reaction is "(G)^p(H)^q / (E)^m(F)^n". The equilibrium constant is a measure of the extent to which a chemical reaction proceeds towards the formation of products at equilibrium.
Ajụjụ 4 Ripọtì
When platinum electrodes are used during the electrolysis of copper (ll) tetraoxosulphate (lV) solution, the solution gets progressively?
Akọwa Nkọwa
Ajụjụ 5 Ripọtì
Which of the additives could improve the quality of steel?
Ajụjụ 6 Ripọtì
A mixture of 0.20 mole of Ar, 0.20 mole of N2 and 0.30 mole of He exerts a total pressure of 2.1 atm. The partial pressure of He in the mixture is?
Akọwa Nkọwa
We are given a mixture of Ar, N2, and He gases with their respective mole amounts and total pressure. To find the partial pressure of He, we can use Dalton's Law of Partial Pressures, which states that the total pressure of a mixture of gases is equal to the sum of the partial pressures of the individual gases. First, we need to calculate the total number of moles of gas in the mixture: 0.20 mol Ar + 0.20 mol N2 + 0.30 mol He = 0.70 mol total Next, we can use the following formula to calculate the partial pressure of He: partial pressure of He = (moles of He / total moles of gas) x total pressure partial pressure of He = (0.30 mol / 0.70 mol) x 2.1 atm partial pressure of He = 0.90 atm Therefore, the partial pressure of He in the mixture is 0.90 atm. Answer choice (a) is correct.
Ajụjụ 7 Ripọtì
Use the graph above to answer this question.
A sample, X, solid at room temperature, was melted, heated to a temperature of 358 K and allowed to cool as shown in OPQR. The section PQ indicates that X is
Akọwa Nkọwa
Based on the graph, section PQ represents a solidification process, where the sample X is cooling and changing from a liquid state to a solid state. Since the cooling curve shows a sharp decrease in temperature without any plateau, it indicates that the solidification process is happening at a constant temperature. This suggests that X is a pure compound with a fixed melting and freezing point. Therefore, the correct answer is: "a pure compound."
Ajụjụ 8 Ripọtì
A saturated solution of AgCl was found to have a concentration of 1.30 x 10-5 mol dm-3. The solubility product of AgCl therefore is?
Akọwa Nkọwa
The concentration of a saturated solution of AgCl is given as 1.30 x 10^-5 mol/dm^3. We can use this information to calculate the solubility product of AgCl. The solubility product (Ksp) of a sparingly soluble salt is the product of the concentrations of its ions in a saturated solution, with each concentration raised to a power equal to the number of ions produced by one formula unit of the salt. For AgCl, the dissociation reaction is: AgCl(s) ⇌ Ag+(aq) + Cl-(aq) The concentration of Ag+ and Cl- in a saturated solution of AgCl is the same and can be represented as x mol/dm^3. Therefore, the Ksp for AgCl can be calculated as follows: Ksp = [Ag+][Cl-] = x * x = x^2 Since the concentration of Ag+ and Cl- in a saturated solution of AgCl is the same, we can assume that x is equal to the given concentration of 1.30 x 10^-5 mol/dm^3. Therefore, Ksp = (1.30 x 10^-5)^2 = 1.69 x 10^-10 mol^2/dm^6. So, the correct answer is option (C): 1.69 x 10^-10 mol^2/dm^6.
Ajụjụ 9 Ripọtì
Choose the correct answer from the options above.
When Fehling's solution is added to two isomeric carbonyl compounds X and Y with the molecular formula C 5H 10O, compound X gives a red precipitate while Y does not react. It can be inferred that X is
Ajụjụ 10 Ripọtì
In the periodic table, what is the property that decreases along the periodic and increases down the group?
Akọwa Nkọwa
The property that decreases along the periodic table and increases down the group is the atomic radius. The atomic radius is the distance between the nucleus and the outermost electrons of an atom. As we move from left to right across a period, the number of protons and electrons increases, and the size of the atom decreases due to the stronger attraction between the positively charged nucleus and the negatively charged electrons. As we move down a group, the number of energy levels increases, resulting in a larger size of the atom. Therefore, the atomic radius decreases along the periodic table and increases down the group.
Ajụjụ 11 Ripọtì
25 cm3 of 0.02 M KOH neutralized 0.03 g of a monobasic organic acid having the general formula CnH2n + 1 COOH. The molecular formula of the acid is?(C = 12, H = 1, O = 16)
Akọwa Nkọwa
Conc. of KOH = 39 x 0.02 = 0.78 g/dm3
25cm3 require 0.03 g of CnH2n + 1 COOH 1000CM3 → x x = (1000) / (25) x 0.03 = 1.2g
CnH2n + 1COOH + KOH → CnH2n + 1 COOK + H2O
from (MAVA) / (MB x VB) = (1) / (1) (MA x 1.2) / (0.02 x 0.78) = (1) / (1)
MA = 0.013
Con. = Molarity x Molecular Mass
Molecular Mass = (0.78) / (0.013) x 60
CnH2n + 1 COOH = 60
12n + 1 (2n + 1) + 12 + 32 + 1 = 6
14n = 60 - 46; n = (14) / (14) = 1
∴ CH3COOH
Ajụjụ 12 Ripọtì
If 30cm3 of oxygen diffuses through porous plug in 7s, how long will it take 60 cm3 of chlorine to diffuse through the same plug?
Akọwa Nkọwa
(R)o/(R)a √(M)a/(MO (30/7)/(60/t) = √ (35.5)/(16)
(60/t) (30/7)/ √(35.5)/(16) t = (60/30)/(7) x (4)√35.5
t = (60 x 7 x √35.5)/(30 x 4) = 21s
Ajụjụ 14 Ripọtì
The reaction of an alkanol with an alkanoic acid in the presence of concentrated H2SO4 will produce an?
Akọwa Nkọwa
The reaction described involves the reaction of an alkanol with an alkanoic acid in the presence of concentrated H2SO4. This is a classic example of an esterification reaction. Esterification reactions usually yield an ester as the product. Therefore, the answer to this question is an "alkanoate" because an ester is also known as an alkanoate.
Ajụjụ 15 Ripọtì
Oxygen is a mixture of two isotopes O and O with relative abundances of 90% and 10% respectively. The relative atomic mass of oxygen is?
Akọwa Nkọwa
The relative atomic mass of oxygen is the weighted average of the masses of its isotopes, taking into account their relative abundances. So, if oxygen has two isotopes, O and O, with relative abundances of 90% and 10% respectively, we can calculate its relative atomic mass as follows: Relative Atomic Mass = (mass of O x relative abundance of O) + (mass of O x relative abundance of O) We can find the masses of each isotope of oxygen in the periodic table. O has a mass of 16 atomic mass units (amu), while O has a mass of 18 amu. Substituting the values into the equation, we get: Relative Atomic Mass of oxygen = (16 amu x 90%) + (18 amu x 10%) = (14.4 amu) + (1.8 amu) = 16.2 amu Therefore, the relative atomic mass of oxygen is 16.2 amu. is the correct answer.
Ajụjụ 16 Ripọtì
(1 /(2) N2(g) + (1) / (2)O2(g) → NO(g) ∆H° 89 KJ mol-. If the entropy change for the reaction above at 25°C is 11.8 J mol-, calculate the change in free energy. ∆G°, for the reaction at 25°C?
Akọwa Nkọwa
∆G = ∆H - T∆S; T = 2s° + 273 = 298 k
∆G = 89 - 298 x 11.8 = -3427.4
Ajụjụ 17 Ripọtì
An element, X, forms a volatile hydride XH3 with a vapour density of 17.0. The relative atomic mass of X is?
(H = 1)
Akọwa Nkọwa
Molecular Mass = Vapour density x 2 = 17.0 x 2 = 34.0
∴ Relative Atomic Mass of X in XH3 = 34 - 3
= 31.0 ∴ x = 31.0
Ajụjụ 18 Ripọtì
One method of driving the position of equilibrium of an endothermic reaction forward is to?
Akọwa Nkọwa
An endothermic reaction is a reaction that absorbs heat from the surroundings. To drive the position of equilibrium of an endothermic reaction forward, we need to provide more heat to the reaction system. Therefore, increasing the temperature at constant pressure is the correct option as it provides the necessary activation energy to the reactants and helps to overcome the energy barrier required to break the bonds and form products. This leads to an increase in the rate of the forward reaction and ultimately shifts the equilibrium position in favor of the products. Conversely, decreasing the temperature at constant pressure would result in a decrease in the rate of the forward reaction and would shift the equilibrium position towards the reactants.
Ajụjụ 19 Ripọtì
Which of the following gases is the most dangerous pollutant?
Akọwa Nkọwa
Carbon monoxide is a colourless, odourless and tasteless gas that is produced by the incomplete combustion of fossil fuel. It is the most dangerous of all the pollutant gases and can cause death in very high concentrations.
Ajụjụ 20 Ripọtì
The temperature of a body decreases when drops of liquid placed on it evaporate because?
Akọwa Nkọwa
When drops of liquid are placed on a body, they will evaporate if the body temperature is higher than the boiling point of the liquid. Evaporation is a cooling process that occurs when molecules escape from the surface of a liquid and take away heat energy with them. In the case of drops of liquid on a body, the heat energy required for the phase transition from liquid to gas is drawn from the body, causing it to cool down. This is known as the heat of vaporization, and it is the reason why the temperature of a body decreases when drops of liquid placed on it evaporate. Therefore, the correct answer to the question is "the heat of vaporization is drawn from the body causing it to cool".
Ajụjụ 21 Ripọtì
If the rate law obtained for a given reaction is rate = k(X)n (Y)m, what is the overall order of the reaction?
Akọwa Nkọwa
The rate law of a chemical reaction describes the relationship between the rate of the reaction and the concentrations of the reactants. The rate law equation given is: rate = k(X)^n(Y)^m where k is the rate constant, (X) and (Y) are the concentrations of the reactants X and Y, and n and m are the reaction orders with respect to X and Y, respectively. The overall order of the reaction is the sum of the reaction orders with respect to each reactant. In this case, the overall order of the reaction is: n + m Therefore, the correct answer is "n + m". To summarize, the overall order of the reaction is the sum of the reaction orders with respect to each reactant, and it is given by the exponents of the concentrations in the rate law equation.
Ajụjụ 22 Ripọtì
A compound that will NOT produce oxygen on heating is?
Ajụjụ 23 Ripọtì
\(2\mathrm{H_2S(g)} + \mathrm{SO_2(g)} + \mathrm{H_2O(l)} \rightarrow 3\mathrm{S(s)} + 3\mathrm{H_2O(l)}\)..........(I)
\(3\mathrm{CuO(s)} + 2\mathrm{NH_3(g)} \rightarrow 3\mathrm{Cu(s)} + 3\mathrm{H_2O(l)} + \mathrm{N_2(g)}\).........(II)
In the equations above, the oxidizing agent (I) and the reducing agent in (II) respectively are?
Ajụjụ 24 Ripọtì
The synthetic detergents are preferred to soap for laundry using hard water because?
Ajụjụ 25 Ripọtì
Sodium hydroxide is prepared commercially from sodium chloride solution by?
Akọwa Nkọwa
Ajụjụ 26 Ripọtì
2SO2(g) + O2(g) 2SO3. In the reaction above, the standard heats of formation of SO2(g) and SO3(g) are -297 KJ mol-1 respectively. The heat change of the reaction is?
Akọwa Nkọwa
2SO2 + O2 → 2SO3
2(-297) 2(-396)
-594 -792
Heat change = heat of product - reactants = -792 - (-594) = 198kj mol-
Ajụjụ 28 Ripọtì
On recrystallization, 20g of magnesium tetraoxosuphate (IV) forms 41 g of magnesium tetraoxosulphate (IV) crystals, MgSO4. yH2O. The value of y is?(Mg = 24, S = 32, O = 16, H = 1)
Akọwa Nkọwa
MgSO4. YH2O
120g (MgSo4) contains - 18g (water)
20g contain (41 - 20g) water.
120g → 18 y; 20 → 21
∴ 18 y = (126)/(18) = 7
Ajụjụ 29 Ripọtì
The compound above is the product of the oxidation of
Ajụjụ 30 Ripọtì
2X(aq) + MnO2(s) + 4H+(aq) → X2(g) + Mn2+(aq) + 2H2O.
The reaction above can be used for the laboratory preparation of all halogens except fluorine because it is?
Ajụjụ 31 Ripọtì
Synthetic rubber obtained by the polymerization of chlorobutadiene in the presence of sodium is called?
Akọwa Nkọwa
The synthetic rubber obtained by the polymerization of chlorobutadiene in the presence of sodium is called neoprene. Neoprene is a type of synthetic rubber that is known for its resistance to oil, chemicals, and heat. It was first developed in the early 1930s and has since been used in a variety of applications, including wetsuits, orthopedic braces, electrical insulation, and automotive fan belts.
Ajụjụ 32 Ripọtì
An undesirable paraffin in the petroleum industry which is particularly prone to knocking is?
Akọwa Nkọwa
The undesirable paraffin in the petroleum industry that is particularly prone to knocking is n-heptane. Knocking refers to the uncontrolled explosion of the air/fuel mixture in the engine's combustion chamber, which can cause significant damage. The octane rating is used to indicate the anti-knock quality of fuels, with iso-octane assigned a rating of 100, and n-heptane assigned a rating of 0. Fuels with a higher octane rating are less likely to knock. Therefore, the presence of n-heptane in gasoline can reduce its octane rating and increase the likelihood of engine knocking.
Ajụjụ 33 Ripọtì
The removal of rust from iron by treatment with tetraoxosulphate (IV) acid is based on the?
Ajụjụ 34 Ripọtì
Which of the aqueous solutions with the pH values below will liberate hydrogen when it react with magnesium metals?
Akọwa Nkọwa
The aqueous solution with a pH of 3.0 will liberate hydrogen when it reacts with magnesium metal. The liberation of hydrogen occurs because the solution with a pH of 3.0 is acidic. Acids are substances that donate protons (H+) when dissolved in water. When magnesium metal reacts with an acid, such as hydrochloric acid (HCl) or sulfuric acid (H2SO4), hydrogen gas (H2) is produced. In this case, the acidic solution with a pH of 3.0 contains hydrogen ions (H+) that can react with the magnesium metal to produce hydrogen gas (H2) according to the following reaction: Mg(s) + 2H+(aq) → Mg2+(aq) + H2(g) The other solutions with pH values of 6.5, 7.0, and 13.0 are not acidic enough to react with magnesium metal and produce hydrogen gas. A neutral solution with a pH of 7.0 contains an equal concentration of hydrogen ions (H+) and hydroxide ions (OH-), and therefore will not donate enough protons to react with the magnesium metal. A basic solution with a pH of 13.0 contains a high concentration of hydroxide ions (OH-), which will not react with magnesium metal to produce hydrogen gas. It's important to note that the reaction between magnesium metal and an acid can be dangerous and should be conducted with caution under appropriate safety measures.
Ajụjụ 35 Ripọtì
Oxidation of concentrated hydrochloric acid with manganese (IV) oxide liberates a gas used in the?
Akọwa Nkọwa
The oxidation of concentrated hydrochloric acid with manganese (IV) oxide produces a gas called chlorine. Chlorine is a chemical element with the symbol Cl and atomic number 17. It is a yellow-green gas with a pungent and suffocating odor. Chlorine has many uses, such as in the production of various chemicals, water treatment, and sterilization. Therefore, the correct answer to the question is "sterilization of water." Chlorine is commonly used to disinfect water and kill harmful microorganisms, making it safe for human consumption.
Ajụjụ 36 Ripọtì
200 cm3 of air was passed over heated copper in a syringe several times to produce copper (II) oxide.
When cooled, the final volume of air recorded was 158cm3. Estimate the percentage of oxygen in the air?
Akọwa Nkọwa
To estimate the percentage of oxygen in the air, we need to use the fact that copper (II) oxide is formed by the reaction of copper and oxygen in the air. The volume of air decreases as the oxygen is used up in the reaction, so the difference between the initial and final volume of air can be used to estimate the volume of oxygen that reacted with the copper. First, let's calculate the difference in volume of air before and after the reaction: Initial volume of air = 200 cm³ Final volume of air = 158 cm³ Difference in volume = 200 cm³ - 158 cm³ = 42 cm³ Next, we need to assume that the copper reacted completely with the oxygen in the air, which means that all the oxygen in the initial volume of air is accounted for in the difference in volume. This allows us to calculate the percentage of oxygen in the air as follows: Volume of oxygen = Difference in volume = 42 cm³ Total volume of air = Initial volume of air = 200 cm³ Percentage of oxygen in the air = (Volume of oxygen / Total volume of air) x 100 = (42 cm³ / 200 cm³) x 100 = 21% Therefore, the percentage of oxygen in the air is estimated to be 21%.
Ajụjụ 37 Ripọtì
Given that 15.00 cm3 of H2
SO4 was required to completely neutralize 25.00 cm3 of 0.125 mol dm-3 NaOH, calculate the molar concentration of the acid solution:
Akọwa Nkọwa
H2SO4 + 2NaOH → Na2SO4 + 2H2O
(MaVa)/(MBVB) (1)/(2) (MA x 15)/(0.125 x25) = 1/2
MA = 0.104
Ajụjụ 38 Ripọtì
Two elements, P and Q with atomic numbers 11 and 8 respectively , combine chemically to form the compound PxQY. The respective values of x and y are?
Ajụjụ 40 Ripọtì
How many more isomers of the compound above can be obtained ?
Ajụjụ 41 Ripọtì
The hydroxyl ion concentrated, (OH-), in a solution of sodium hydroxide of pH 10.0 is?
Akọwa Nkọwa
Ajụjụ 42 Ripọtì
35 cm3 of hydrogen was sparked with 12 cm3 of oxygen at 110°C and 760mm Hg to produce steam. What percentage of the total volume of gas left after the reaction is hydrogen?
Ajụjụ 43 Ripọtì
How many Faraday of electricity are required to deposit 0.20 mole of nickel, if 0.10 Faraday of electricity deposited 2.98g of nickel during electrolysis of its aqueous solution?(Ni = 58.7, 1F = 96 500 C mol-)
Akọwa Nkọwa
We can use the equation: moles of substance = electric charge (in Coulombs) / (Faraday's constant x charge per ion) We are given that 0.10 Faraday of electricity deposited 2.98g of nickel. The molar mass of nickel is 58.7 g/mol, so 2.98g is equivalent to 0.0508 moles of nickel: moles of Ni = 2.98g Ni / 58.7 g/mol = 0.0508 moles of Ni Now, we can use the equation above to find out how many Faradays are required to deposit 0.20 moles of nickel: 0.20 moles of Ni = electric charge (in Coulombs) / (96,500 C/mol x 2) electric charge (in Coulombs) = 0.20 moles of Ni x 96,500 C/mol x 2 electric charge (in Coulombs) = 38,600 C Therefore, we need 0.40 Faraday of electricity to deposit 0.20 moles of nickel: 0.40 F = 38,600 C / 96,500 C/mol So the answer is (C) 0.40.
Ajụjụ 44 Ripọtì
A sample of a substance containing only C and H burns in excess O2 to yield 4.4g of CO2
And 2.7 g of H2O. The empirical formula of a substance is?(C = 12, O = 16, H = 1)
Akọwa Nkọwa
To determine the empirical formula of the substance, we need to find the ratio of the number of atoms of each element in the compound. First, we need to find the number of moles of each product formed. Using the molar masses of CO2 (44 g/mol) and H2O (18 g/mol), we can calculate: moles of CO2 = 4.4 g / 44 g/mol = 0.1 mol CO2 moles of H2O = 2.7 g / 18 g/mol = 0.15 mol H2O Next, we need to determine the number of moles of carbon and hydrogen that were present in the original sample. We can use the amount of CO2 and H2O produced to calculate this: moles of C = moles of CO2 = 0.1 mol C moles of H = moles of H2O x 2 = 0.3 mol H Now we can find the simplest whole-number ratio of carbon to hydrogen in the compound. Dividing the number of moles of each element by the smallest number of moles gives: C: 0.1 mol / 0.1 mol = 1 H: 0.3 mol / 0.1 mol = 3 So the empirical formula of the compound is CH3. Therefore, the correct option is (a) CH3.
Ajụjụ 45 Ripọtì
The electron configuration of two elements with similar chemical properties are represented by?
Ajụjụ 46 Ripọtì
A major process involved in the softening of hard water is the?
Akọwa Nkọwa
The process involved in the softening of hard water is the conversion of a soluble calcium salt to its trioxocarbonate (IV) through a chemical reaction known as precipitation. Hard water is water that contains high levels of dissolved calcium and magnesium ions. These ions react with soap, making it difficult to lather and causing scale buildup in pipes and appliances. To soften hard water, we need to remove the calcium and magnesium ions from the water. One way to do this is through precipitation. In this process, we add a chemical called a precipitating agent to the hard water. The precipitating agent reacts with the calcium ions, causing them to form a solid precipitate of calcium trioxocarbonate (IV), which can be filtered out of the water. The equation for the reaction is: Ca2+ (aq) + CO32- (aq) → CaCO3 (s) where Ca2+ represents the calcium ions in the water, CO32- represents the carbonate ions in the precipitating agent, and CaCO3 represents the solid precipitate of calcium trioxocarbonate (IV). Once the calcium and magnesium ions have been removed from the water, it is considered soft and is less likely to cause scale buildup in pipes and appliances. Therefore, the correct answer is option (i): conversion of a soluble calcium salt to its trioxocarbonate (IV).
Ajụjụ 47 Ripọtì
2.85g of an oxide of copper gave 2.52g of copper on reduction and 1.90g of another oxide gave 1.52g of copper on reduction. The data above illustrates the law of?
Akọwa Nkọwa
The data given in the question illustrates the law of multiple proportions. This law states that when two elements combine to form two or more compounds, the mass of one element that combines with a fixed mass of the other element are in a ratio of small whole numbers. In the given question, we have two different oxides of copper, and the mass of copper produced on the reduction of these oxides shows a ratio of small whole numbers. Specifically, the ratio of the masses of copper produced is 1.66:1, which can be simplified to 5:3. This indicates that the two oxides have a different ratio of copper to oxygen, as predicted by the law of multiple proportions.
Ajụjụ 48 Ripọtì
The compound above contains
Akọwa Nkọwa
Without knowing the specific compound being referred to, it is impossible to provide a definitive answer. However, based solely on the information given in the question, the possible hybridization states of the carbon atoms in the compound can be determined. The notation used in the options refers to the hybridization state of the carbon atoms, which is determined by the type and number of orbitals used by the carbon atom to form bonds. For example, "Sp3" hybridization indicates that the carbon atom is using one s orbital and three p orbitals to form four sp3 hybrid orbitals for bonding. Similarly, "p2" hybridization indicates that the carbon atom is using two p orbitals for bonding. Therefore, based on the options given, the answer would be either (A) Sp3 hybridized carbon atoms only, (B) p2 hybridized carbon atoms only, (C) a combination of Sp3 and Sp hybridized carbon atoms, or (D) a combination of Sp3 and Sp2 hybridized carbon atoms.
Ajụjụ 49 Ripọtì
What is the oxidation number of Z in K3ZCl6?
Akọwa Nkọwa
The oxidation number of an element in a compound is the charge it would have if all the electrons in the bonds belonged to the more electronegative atom. In this case, we can start by finding the oxidation number of Cl since its value is more common and well-known. The compound is K3ZCl6, which means that the total charge of the compound must be zero since it is not an ion. We know that the potassium ion (K+) has a charge of +1, and there are three of them, so the total charge from potassium is +3. The chloride ion (Cl-) has a charge of -1, and there are six of them, so the total charge from chloride is -6. Therefore, the oxidation number of Z can be found by solving the equation: (+3) + (oxidation number of Z) + (-6) = 0 Solving this equation gives us the oxidation number of Z, which is +3. Therefore, the correct answer is option (B) +3.
