
Assertion: \[\text{BiC}{{\text{l}}_{\text{5}}}\] does not exist.
Reason: In $ \text{Bi} $ inert pair effect is predominant.
Read the above assertion and reason and choose the correct option regarding it.
(A) Both assertion and reason are correct and the reason is the correct explanation of the assertion.
(B) Both assertion and reason are correct but the reason is not the correct explanation of the assertion.
(C) Assertion is correct but the reason is incorrect.
(D) Assertion is incorrect and the reasons are correct.
Answer
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Hint: The inert pair effect is shown by some of the elements of the sixth period wherein the s-electrons in the 6s subshell do not show any tendency to involve in covalent bonding due to the poor shielding effect of the 5d and 4f orbitals present before it.
Complete step by step solution:
Bismuth belongs to the group 15 and the sixth period of the periodic table and it has the electronic configuration $ \left[ \text{Xe} \right]\text{4}{{\text{f}}^{\text{14}}}\text{5}{{\text{d}}^{\text{10}}}\text{6}{{\text{s}}^{\text{2}}}\text{6}{{\text{p}}^{\text{3}}} $ . Due to the presence of the 4f and the 5d orbitals prior to the 6s orbital, their diffuse shape offer poor shielding effect to the electrons of the 6s orbital which being an “s” orbital has higher penetration power than the 4f and the 5d orbital electrons. Hence these electrons are attracted more by the nucleus and they refrain from participating in chemical bonding because the amount of energy required to unpair them and involve the s-orbitals in bond formation will not be recovered by the formation of two extra bonds.
Hence, bismuth forms trichloride involving the three “p” orbitals but does not form the pentachloride involving both the s and p orbitals.
So both the assertion and reason are correct and the reason is the correct explanation of the assertion.
The correct answer is option A.
Note:
The inert pair effect is also shown by two other elements besides bismuth which are thallium from group 13 and lead from group 14, both of them belonging to the same period. This effect is so pronounced for the 6th period due to the diffuse shape of the 4d and 5f levels.
Complete step by step solution:
Bismuth belongs to the group 15 and the sixth period of the periodic table and it has the electronic configuration $ \left[ \text{Xe} \right]\text{4}{{\text{f}}^{\text{14}}}\text{5}{{\text{d}}^{\text{10}}}\text{6}{{\text{s}}^{\text{2}}}\text{6}{{\text{p}}^{\text{3}}} $ . Due to the presence of the 4f and the 5d orbitals prior to the 6s orbital, their diffuse shape offer poor shielding effect to the electrons of the 6s orbital which being an “s” orbital has higher penetration power than the 4f and the 5d orbital electrons. Hence these electrons are attracted more by the nucleus and they refrain from participating in chemical bonding because the amount of energy required to unpair them and involve the s-orbitals in bond formation will not be recovered by the formation of two extra bonds.
Hence, bismuth forms trichloride involving the three “p” orbitals but does not form the pentachloride involving both the s and p orbitals.
So both the assertion and reason are correct and the reason is the correct explanation of the assertion.
The correct answer is option A.
Note:
The inert pair effect is also shown by two other elements besides bismuth which are thallium from group 13 and lead from group 14, both of them belonging to the same period. This effect is so pronounced for the 6th period due to the diffuse shape of the 4d and 5f levels.
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