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Question
Explain the following:
A Gallium has higher ionization enthalpy than Aluminium.
B Boron does not exist as \(B^{3+}\) ion
C Aluminium forms \([AlF_6]^{3-}\)- ion but Boron does not form \([BF_6]^{3-}\)- ion.
D \(PbX_2\) is more stable than \(PbX_4\).
E \(Pb^{4+}\) acts as an oxidizing agent but \(Sn^{2+}\) acts as a reducing agent.
F Electron gain enthalpy of chlorine is more negative as compared to fluorine.
G \(Tl (NO_3)_3\) acts as an oxidizing agent.
H Carbon shows catenation property but lead does not.
I \(BF_3\) does not hydrolyze.
J Why does the element silicon not form a graphite-like structure whereas carbon does.
A Gallium has higher ionization enthalpy than Aluminium.
B Boron does not exist as \(B^{3+}\) ion
C Aluminium forms \([AlF_6]^{3-}\)- ion but Boron does not form \([BF_6]^{3-}\)- ion.
D \(PbX_2\) is more stable than \(PbX_4\).
E \(Pb^{4+}\) acts as an oxidizing agent but \(Sn^{2+}\) acts as a reducing agent.
F Electron gain enthalpy of chlorine is more negative as compared to fluorine.
G \(Tl (NO_3)_3\) acts as an oxidizing agent.
H Carbon shows catenation property but lead does not.
I \(BF_3\) does not hydrolyze.
J Why does the element silicon not form a graphite-like structure whereas carbon does.
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Solution
A. In \(Ga\), due to ineffective shielding of valence electrons by the \(3d\) electrons, the effective nuclear charge increases and the electrons are much more attracted to the nucleus than in \(Al\).
Hence, due to higher value of effective nuclear charge in gallium, it has higher ionization enthalpy than \(Al\).
B.Boron is the smallest element of the group.
Due to its smallest size, high electronegativity and a high sum of the first three ionization enthalpies to remove all the three valence electrons, it is highly unlikely to form \(B^{3+}\) ion.
Hence, \(B\) does not exist as \(B^{3+}\) ion.
C.\(B\) has \(2^{nd}\) shell as its valence shell, which does not have vacant \(d\)-orbital. Therefore, Boron cannot form more bonds than four.
But \(Al\) has empty \(3d\)-orbitals to accommodate the electrons from \(F\)- ions to form \([AlF_6]^{3-}\) ion.
D. \(Pb\) is the heaviest element of group \(14\).
Due to inert pair effect the stability of \(+4\) oxidation state decreases down the group and stability of \(+2\) increases down the group, due to non availability of \(ns\) electrons (where \(n\) is the valence shell) for bonding.
Hence \(PbX_2\) is more stable than \(PbX_4\) as \(Pb\) is more stable in \(+2\) than in \(+4\) state.
E. \(Sn\) and \(Pb\) are the lower member of group \(14\)
i.e., they lie at the bottom. Group \(14\) elements have general electronic configuration of \(ns^2 np^2\) i.e., \(4\) electrons are available for bonding.
Due to the inert pair effect, stability of \(+4\) state decreases down the group.
It is more prominent in \(Pb\) than in \(Sn\). \(+2\) is more stable oxidation state of \(Pb\) than \(+4\).
\(Pb^{4+}\) tries to get reduced to \(+2\) state by accepting electrons and oxidizing other species.
\(Sn^{2+}\) tries to oxidize to \(Sn^{4+}\) by donating electrons and reducing other species.
F. \(F\) has very small size as compared to \(Cl\).
The added electron in the \(2p\) orbital of \(F\) experiences more interelectronic repulsion than the added electron in the \(3p\) orbital of chlorine.
This is because the size of \(3p\) is greater than the size of \(2p\) orbital.
Therefore, it is not easier to add electron to \(F\) than in \(Cl\).
Hence, \(Cl\) has a higher negative value of electron gain enthalpy than \(F\).
G. In \(Tl(NO_3)_3\), the oxidation state of \(Tl\) is \(+3\).
Due to the inert pair effect in group \(13\), \(Tl\) is more stable in \(+1\) state than in \(+3\) state.
Therefore, \(Tl(NO_3)_3\) act as oxidizing agent i.e., it oxidizes other species by gaining \(2\) electrons from them and itself reduces to a more stable state of \(+1\).
H. Down the group, the size of atoms increases and cannot form a stable \(M\)-\(M\) bond.
\(C\)-\(C\) bond is stronger and can effectively form \(p \pi\)−\(p \pi\) bonds with itself and therefore shows the property of catenation.
\(Pb\) due to large size and less electronegativity is not able to form \(Pb\)-\(Pb\) strong bonds and hence does not show catenation.
I. \(BF_3\) does not get hydrolyzed completely.
Instead, it hydrolyzes incompletely to form boric acid and fluoroboric acid. This is because the \(HF\) first formed reacts with \(H_3BO_3\). These equations can be written as:
\(4BF_3 + 12H_2O \longrightarrow 4H_3BO_3 + 12HF\)
\(3H_3BO_3 + 12 HF \longrightarrow 3H^+ + 3[BF_4]^- + 9H_2O\)
\(4BF_3 + 3H_2O \longrightarrow 4H_3BO_3 + 3 [BF_4]^- + 3H^+\)Thus, \(BF_3\) does not hydrolyze completely.
J. Graphite has a layered structure where each layer is composed of planar hexagonal rings of carbon atoms.
\(C\)-\(C\) bond is stronger and can show higher catenation tendency than \(Si\).
Down the group, the tendency to show catenation decreases due to larger size of atoms \(Sn\) and less electronegativity of elements like \(Sn\), \(Ge\).
Thus, \(Si\) shows very less catenation and cannot form rings like carbon atoms in graphite.
Hence, due to higher value of effective nuclear charge in gallium, it has higher ionization enthalpy than \(Al\).
B.Boron is the smallest element of the group.
Due to its smallest size, high electronegativity and a high sum of the first three ionization enthalpies to remove all the three valence electrons, it is highly unlikely to form \(B^{3+}\) ion.
Hence, \(B\) does not exist as \(B^{3+}\) ion.
C.\(B\) has \(2^{nd}\) shell as its valence shell, which does not have vacant \(d\)-orbital. Therefore, Boron cannot form more bonds than four.
But \(Al\) has empty \(3d\)-orbitals to accommodate the electrons from \(F\)- ions to form \([AlF_6]^{3-}\) ion.
D. \(Pb\) is the heaviest element of group \(14\).
Due to inert pair effect the stability of \(+4\) oxidation state decreases down the group and stability of \(+2\) increases down the group, due to non availability of \(ns\) electrons (where \(n\) is the valence shell) for bonding.
Hence \(PbX_2\) is more stable than \(PbX_4\) as \(Pb\) is more stable in \(+2\) than in \(+4\) state.
E. \(Sn\) and \(Pb\) are the lower member of group \(14\)
i.e., they lie at the bottom. Group \(14\) elements have general electronic configuration of \(ns^2 np^2\) i.e., \(4\) electrons are available for bonding.
Due to the inert pair effect, stability of \(+4\) state decreases down the group.
It is more prominent in \(Pb\) than in \(Sn\). \(+2\) is more stable oxidation state of \(Pb\) than \(+4\).
\(Pb^{4+}\) tries to get reduced to \(+2\) state by accepting electrons and oxidizing other species.
\(Sn^{2+}\) tries to oxidize to \(Sn^{4+}\) by donating electrons and reducing other species.
F. \(F\) has very small size as compared to \(Cl\).
The added electron in the \(2p\) orbital of \(F\) experiences more interelectronic repulsion than the added electron in the \(3p\) orbital of chlorine.
This is because the size of \(3p\) is greater than the size of \(2p\) orbital.
Therefore, it is not easier to add electron to \(F\) than in \(Cl\).
Hence, \(Cl\) has a higher negative value of electron gain enthalpy than \(F\).
G. In \(Tl(NO_3)_3\), the oxidation state of \(Tl\) is \(+3\).
Due to the inert pair effect in group \(13\), \(Tl\) is more stable in \(+1\) state than in \(+3\) state.
Therefore, \(Tl(NO_3)_3\) act as oxidizing agent i.e., it oxidizes other species by gaining \(2\) electrons from them and itself reduces to a more stable state of \(+1\).
H. Down the group, the size of atoms increases and cannot form a stable \(M\)-\(M\) bond.
\(C\)-\(C\) bond is stronger and can effectively form \(p \pi\)−\(p \pi\) bonds with itself and therefore shows the property of catenation.
\(Pb\) due to large size and less electronegativity is not able to form \(Pb\)-\(Pb\) strong bonds and hence does not show catenation.
I. \(BF_3\) does not get hydrolyzed completely.
Instead, it hydrolyzes incompletely to form boric acid and fluoroboric acid. This is because the \(HF\) first formed reacts with \(H_3BO_3\). These equations can be written as:
\(4BF_3 + 12H_2O \longrightarrow 4H_3BO_3 + 12HF\)
\(3H_3BO_3 + 12 HF \longrightarrow 3H^+ + 3[BF_4]^- + 9H_2O\)
\(4BF_3 + 3H_2O \longrightarrow 4H_3BO_3 + 3 [BF_4]^- + 3H^+\)Thus, \(BF_3\) does not hydrolyze completely.
J. Graphite has a layered structure where each layer is composed of planar hexagonal rings of carbon atoms.
\(C\)-\(C\) bond is stronger and can show higher catenation tendency than \(Si\).
Down the group, the tendency to show catenation decreases due to larger size of atoms \(Sn\) and less electronegativity of elements like \(Sn\), \(Ge\).
Thus, \(Si\) shows very less catenation and cannot form rings like carbon atoms in graphite.
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