(a) Atoms of four non-metallic elements in the same group of the periodic table are arranged in order of increasing atomic radius as R < T < W < X. Which of the elements
(i) would readily lose electron(s) from the outermost shell;
(ii) is most electronegative;
(iii) would T displace from aqueous solution;
(iv) is at the top of the group?
(b) The following table shows the electronic configuration of two elements Y and Z.
| Element |
Electronic Configuration
|
| Y |
1s\(^2\) 2s\(^2\) 2p\(^5\) |
| Z |
1s\(^2\) 2s\(^2\) 2p\(^5\) 3s\(^1\) |
(i) Name the I. group to which Y belongs II. period to which Z belongs.
(ii) What is the number of protons present in an atom of Z?
(iii) How many unpaired electrons are in an atom of Y?
(iv) Write the formula of the compound formed between Y and Z
(c) Name the type of bond(s) that exist(s) in each of the following compounds.
(i) CaCl\(_2\); (ii) NH\(_4\)Cl; CCl\(_4\)
(d) Describe how the conductance of a molar solution of ammonia compares to that of sodium hydroxide solution.
(e) State the type of reaction illustrated by each of the following equations:
(i) CH\(_3\)CH\(_2\)OH\(_{(/)}\) + CH\(_3\)COOH\(_{(aq)}\) \(\rightleftharpoons\) CH\(_3\)COOC\(_2\)H\(_5\)\(_{(/)}\) + H\(_2\)O\(_{(/)}\)
(ii) H+\(_{Y^+_{(aq)}\) + OH\(^+_{(aq)}\) \(\to\) H\(_2\)O
(f) Determine the volume of the residual gas when 20.0cm\(^3\) of hydrogen was sparked with 15.0cm\(^3\) of oxygen and the resulting mixture cooled to room temperature.
(a) Four non-metals of the same group with increasing atomic radius \( R < T < W < X \), so R is at the top (smallest) and X at the bottom (largest).
- (i) Would most readily lose electrons from the outermost shell: X (largest atom, weakest hold on outer electrons, least electronegative).
- (ii) Most electronegative: R (smallest atom, at the top).
- (iii) For non-metals reactivity decreases down the group, so T can displace those below it, namely W and X, from aqueous solution.
- (iv) At the top of the group: R.
(b) Y: \( 1s^2\,2s^2\,2p^5 \) (fluorine). Z: \( 1s^2\,2s^2\,2p^6\,3s^1 \) (sodium), giving 11 electrons in the neutral atom.
| Element | Electronic configuration |
|---|
| Y | \( 1s^2\,2s^2\,2p^5 \) |
| Z | \( 1s^2\,2s^2\,2p^6\,3s^1 \) |
- (i) I. Y belongs to Group VII (the halogens). II. Z is in Period 3.
- (ii) Number of protons in an atom of Z \( = 11 \).
- (iii) In Y the \( 2p^5 \) sub-shell has one unpaired electron, so there is 1 unpaired electron.
- (iv) Y forms \( Y^- \) and Z forms \( Z^+ \), so the compound is ZY (that is \( NaF \)).
(c) Bond types
- (i) \( CaCl_2 \): ionic (electrovalent) bond.
- (ii) \( NH_4Cl \): ionic bond (between \( NH_4^+ \) and \( Cl^- \)) together with covalent and coordinate (dative) bonds within the ammonium ion.
- (iii) \( CCl_4 \): covalent bond.
(d) Conductance of ammonia compared with sodium hydroxide
Ammonia solution is a weak base that is only slightly ionized, so it produces few ions and conducts electricity poorly. Sodium hydroxide is a strong base that is fully ionized, producing many ions, so at the same molar concentration it conducts electricity much better than ammonia.
(e) Type of reaction
- (i) \( CH_3CH_2OH + CH_3COOH \rightleftharpoons CH_3COOC_2H_5 + H_2O \): esterification (a reversible condensation reaction).
- (ii) \( H^+ + OH^- \rightarrow H_2O \): neutralization.
(f) Volume of residual gas
\[ 2H_2 + O_2 \rightarrow 2H_2O \]
\( 20.0\ cm^3 \) of \( H_2 \) reacts with \( \dfrac{20.0}{2} = 10.0\ cm^3 \) of \( O_2 \). Oxygen supplied is \( 15.0\ cm^3 \), so oxygen left over \( = 15.0 - 10.0 = 5.0\ cm^3 \). The water formed condenses to a liquid on cooling.
Residual gas = 5.0 cm\(^3\) of oxygen.
(a) Four non-metals of the same group with increasing atomic radius \( R < T < W < X \), so R is at the top (smallest) and X at the bottom (largest).
- (i) Would most readily lose electrons from the outermost shell: X (largest atom, weakest hold on outer electrons, least electronegative).
- (ii) Most electronegative: R (smallest atom, at the top).
- (iii) For non-metals reactivity decreases down the group, so T can displace those below it, namely W and X, from aqueous solution.
- (iv) At the top of the group: R.
(b) Y: \( 1s^2\,2s^2\,2p^5 \) (fluorine). Z: \( 1s^2\,2s^2\,2p^6\,3s^1 \) (sodium), giving 11 electrons in the neutral atom.
| Element | Electronic configuration |
|---|
| Y | \( 1s^2\,2s^2\,2p^5 \) |
| Z | \( 1s^2\,2s^2\,2p^6\,3s^1 \) |
- (i) I. Y belongs to Group VII (the halogens). II. Z is in Period 3.
- (ii) Number of protons in an atom of Z \( = 11 \).
- (iii) In Y the \( 2p^5 \) sub-shell has one unpaired electron, so there is 1 unpaired electron.
- (iv) Y forms \( Y^- \) and Z forms \( Z^+ \), so the compound is ZY (that is \( NaF \)).
(c) Bond types
- (i) \( CaCl_2 \): ionic (electrovalent) bond.
- (ii) \( NH_4Cl \): ionic bond (between \( NH_4^+ \) and \( Cl^- \)) together with covalent and coordinate (dative) bonds within the ammonium ion.
- (iii) \( CCl_4 \): covalent bond.
(d) Conductance of ammonia compared with sodium hydroxide
Ammonia solution is a weak base that is only slightly ionized, so it produces few ions and conducts electricity poorly. Sodium hydroxide is a strong base that is fully ionized, producing many ions, so at the same molar concentration it conducts electricity much better than ammonia.
(e) Type of reaction
- (i) \( CH_3CH_2OH + CH_3COOH \rightleftharpoons CH_3COOC_2H_5 + H_2O \): esterification (a reversible condensation reaction).
- (ii) \( H^+ + OH^- \rightarrow H_2O \): neutralization.
(f) Volume of residual gas
\[ 2H_2 + O_2 \rightarrow 2H_2O \]
\( 20.0\ cm^3 \) of \( H_2 \) reacts with \( \dfrac{20.0}{2} = 10.0\ cm^3 \) of \( O_2 \). Oxygen supplied is \( 15.0\ cm^3 \), so oxygen left over \( = 15.0 - 10.0 = 5.0\ cm^3 \). The water formed condenses to a liquid on cooling.
Residual gas = 5.0 cm\(^3\) of oxygen.
