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JAMB UTME - Chemistry - 2024

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Ajụjụ 1 Ripọtì

The number of geometrical isomers of butene are

2

3

5

7

Akọwa Nkọwa

To understand the geometrical isomers of butene, we need to explore its structure. Butene has four carbon atoms, and there are various structural forms that butene can take. These structural forms include linear or branched chains, with a double bond present between carbon atoms.

Geometric isomerism is a type of stereoisomerism. It occurs due to restricted rotation around the double bond, leading to different spatial arrangements of groups attached to the carbons forming the double bond. The geometric isomerism primarily occurs in alkenes like butene where the positions of substituents can vary.

Let's consider the different types of butene, focusing on the possibility of geometrical isomerism:

1-butene: This is a straight-chain isomer with the double bond starting at the first carbon. It doesn't show geometrical isomerism because the double bond is at the end of the chain, so there are no different spatial arrangements of groups possible.
2-butene: Here the double bond is between the second and third carbon. In this case, geometrical isomerism is possible because different groups (or atoms) attached to the carbon atoms can have different spatial arrangements. Specifically:
- cis-2-butene: Both methyl groups (CH₃) are on the same side of the double bond.
- trans-2-butene: The methyl groups are on opposite sides of the double bond.
Isobutene (or 2-methylpropene): This is a branched isomer, and it does not have geometric isomers because the substituents around the double bond do not allow for different spatial configurations on either side of the double bond.

In conclusion, for butene, only 2-butene has geometrical isomers (cis and trans). Therefore, the number of geometric isomers is 2.

Gụọ akwụkwọ nkuzi na Organic Compounds (JAMB)

Organic Compounds

Ajụjụ 3 Ripọtì

If the solubility of KNO $_{3}$ at 30 $^{0}$ C is 3.10 mol/dm $^{3}$ a solution containing 303g/dm $^{3}$ KNO $_{3}$ is likely to be

saturated

unsaturated

supersaturated

at saturation point

Ajụjụ 4 Ripọtì

Na $_{2}$ X ⇌ 2Na $^{+}$ + X $^{2 -}$

The bond between Na and X is likely to be

covalent

dative

ionic

metallic

Atomic Structure And Bonding

Ajụjụ 12 Ripọtì

H $_{2}$ S $_{(g)}$ + Cl $_{2}$ $_{(g)}$ → 2HCl $_{(g)}$ + S $_{(s)}$

What is the change in oxidation state of sulphur from reactant to product?

-3 to -1

-3 to 1

-2 to 0

-1 to 0

Oxidation And Reduction

Ajụjụ 13 Ripọtì

The hybridization scheme in ethyne is

sp

sp

^{3}

sp

^{2}

d

sp

^{2}

s

Akọwa Nkọwa

Ethyne, also known as acetylene, is a simple alkyne with the chemical formula C₂H₂. In ethyne, each carbon atom is bonded to two other atoms: one hydrogen atom and the other carbon atom. The molecular structure of ethyne is linear, with a triple bond between the two carbon atoms.

To determine the hybridization scheme in ethyne, we need to examine the arrangement of the electron pairs around each carbon atom. In ethyne, each carbon atom is forming two sigma (σ) bonds and two pi (π) bonds. Let's explain:

The two carbon atoms are linked by a triple bond, consisting of one sigma bond and two pi bonds. The sigma bond is formed by the overlap of orbitals directly between the two carbon nuclei. The pi bonds are formed by the overlap of p orbitals above and below the plane of the molecule.
The carbon atoms in ethyne are bonded to only two atoms, which is indicative of a linear geometry.

When we consider the hybridization of the carbon atoms, we focus on the formation of sigma bonds and lone pairs. In ethyne, each carbon atom utilizes two orbitals to form sigma bonds: one with the hydrogen atom and one with the other carbon atom. This implies that each carbon atom in ethyne must use two hybrid orbitals.

The two hybrid orbitals formed by each carbon atom in ethyne are a result of mixing one s orbital with one p orbital. This hybridization is referred to as sp hybridization, characterized by a linear electron geometry. The remaining two unhybridized p orbitals on each carbon atom are responsible for forming the two pi bonds in the triple bond.

In conclusion, the hybridization scheme in ethyne is sp.

Gụọ akwụkwọ nkuzi na Atomic Structure And Bonding (JAMB)

Atomic Structure And Bonding

Ajụjụ 15 Ripọtì

An example of a compound that is acidic in solution is

H

_{3}

PO

_{4}

NH

_{4}

Cl

Na

_{2}

CO

_{3}

MgCl

_{2}

Acids, Bases And Salts

Ajụjụ 19 Ripọtì

The volume in cm $^{3}$ of a 0.12 moldm $^{- 3}$ HCl required to completely neutralize a 20cm $^{3}$ of 0.20 moldm $^{- 3}$ of NaOH is

12.00

0.0012

33.33

2.40

Acids, Bases And Salts

Ajụjụ 20 Ripọtì

$_{7}^{14}$ N + X → $_{6}^{14}$ C + $_{1}^{1}$ P

In the reaction above, X is

a proton

an alpha particle

a beta particle

a neutron

Akọwa Nkọwa

To determine what particle X is, we need to understand the reaction given:

N + X → \^146\\ C + \^11\\P

The notation in nuclear reactions is important. The numbers on top (superscripts) are the mass numbers, which represent the total number of protons and neutrons. The numbers on the bottom (subscripts) are the atomic numbers, which represent the number of protons.

Here's what we have:

N: This is likely to represent some isotope of nitrogen. However, we do not have its specific isotope number, so we represent it with N.
146/6 C: This represents a carbon isotope with a mass number of 146 and 6 protons.
11/1 P: This represents a particle with a mass number of 11 and 1 proton, which is likely a proton since it's positively charged with atomic number 1 and a larger mass compared to typical neutrons.

Let's consider the conservation of mass and charge:

1. **Conservation of Mass Number:** The mass number of the reactants should equal the mass number of the products. If N has a mass number 'a' and X has a mass number 'b', then:

a + b = 146 + 11 = 157

2. **Conservation of Atomic Number:** The total number of protons should also be conserved. If N has an atomic number 'c' and X has an atomic number 'd', then:

c + d = 6 + 1 = 7

To satisfy these rules:

- Option X could be a **neutron**, as neutrons have a mass number of 1 and an atomic number of 0, which means they do not affect the atomic number but contribute to the mass number.

Let's verify:

- Assume X is a neutron with a mass number of 1 and an atomic number of 0, which fits the requirement for conservation of atomic mass:

a + 1 = 157 (where a is from N mass number)
c + 0 = 7 (where c is from N atomic number)

Therefore, X is a neutron because it helps conserve both the mass number and the atomic number in the given nuclear reaction.

Gụọ akwụkwọ nkuzi na Structure Of The Atom (WAEC)

Structure Of The Atom

Ajụjụ 21 Ripọtì

CH $_{3}$ -CH $_{2}$ -OH and CH $_{3}$ -O-CH $_{3}$

The relationship between the two compounds above, is that they are

allotropes

isotopes

isomers

isobars

Organic Compounds

Ajụjụ 23 Ripọtì

Aqueous solution of sodium hydroxide can be used to test for the presence of : I. Ca $^{2 +}$ , II. Zn $^{2 +}$ , III. Cu $^{2 +}$

I only

I and III

I, II and III

I and II only

Acids, Bases And Salts

Ajụjụ 25 Ripọtì

Concentrated sodium chloride solution is electrolyzed using mercury cathode and graphite anode. The products at the anode and the cathode respectively are

hydrogen and chlorine

chlorine and hydrogen

chlorine and sodium

sodium and chlorine

Akọwa Nkọwa

When a concentrated sodium chloride solution is electrolyzed using a mercury cathode and graphite anode, the products are hydrogen gas at the cathode and chlorine gas at the anode

At the anode, 2Cl $^{-}$ → Cl $_{2}$ + 2e $^{-}$

At the cathode, 2H $^{+}$ + 2e $^{-}$ → H $_{2}$

During the electrolysis, hydrogen and chloride ions are removed from solution whereas sodium and hydroxide ions are left behind in solution. This means that sodium hydroxide is also formed during the electrolysis of sodium chloride solution.

Gụọ akwụkwọ nkuzi na Electrolysis (JAMB)

Electrolysis

Ajụjụ 27 Ripọtì

CuO $_{s}$ + H $_{2}$ ( $_{g}$ ) ⇌ Cu $_{s}$ + H $_{2}$ O( $_{g}$ )

In the equation above, the effect of increased pressure on the equilibrium position is that

the equilibrium is shifted to the left

the equilibrium is shifted to the right

there is no effect

more H

_{2}

(g

) is produced

Chemical Equilibra

Ajụjụ 28 Ripọtì

The ions responsible for permanent hardness in water are sulphates of

Fe

^{3 +}

and Mg

^{2 +}

Ca

^{2 +}

and Mg

^{2 +}

Fe

^{2 +}

and Fe

^{3 +}

Na

^{+}

and Ca

^{2 +}

Ajụjụ 34 Ripọtì

The IUPAC nomenclature of the complex K $_{4}$ Fe(CN) $_{6}$ is

potassium hexacyanoferrate(II)

potassium hexacyanoferrate(III)

potassium hexacyanoferrate(IV)

potassium hexacyanoferrate(VI)

Organic Compounds

Ajụjụ 36 Ripọtì

An example of highly unsaturated hydrocarbon is

C

_{2}

H

_{2}

CH

_{4}

C

_{2}

H

_{4}

C

_{4}

H

_{10}

Organic Compounds

Ajụjụ 39 Ripọtì

After breathing in a test tube that contains acidified K $_{2}$ Cr $_{2}$ O $_{7}$ , a man noticed the change in the colour of K $_{2}$ Cr $_{2}$ O $_{7}$ from orange to green. This suggests the presence of

alkanone

alkanal

alkanol

alkanoic acid

Redox Reactions

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