Explanation
The inertness of s subshell electrons towards the bond formation is called inert pair effect or it can be said as the inactiveness of electrons present in outermost shell (i.e. $$ns^2$$) to get unpaired and involve in bond formation is called inert pair effect.
for example:
1) In 13th group, thallium can exhibit $$+1$$ and $$+3$$ oxidation states but it is stable in $$+1$$ oxidation state only due to inert pair effect
2) In 14th group , lead shows both $$+2$$ and $$+4$$ oxidation states but it is stable in $$+2$$ oxidation state due to inert pair effect.
3)In group 15 elements, the stability of $$+3$$ oxidation state increases down the group and that of $$+5$$ oxidation state decreases down the group, due to inert pair effect."
The inertness of s subshell electrons towards the bond formation is called inert pair effect or it can be said as the inactiveness of electrons present in the outermost shell (i.e. $$ns^2$$) to get unpaired and involve in bond formation is called inert pair effect.
2) In 14th group, lead shows both $$+2$$ and $$+4$$ oxidation states but it is stable in $$+2$$ oxidation state due to inert pair effect.
3)In group 15 elements, ($$Bi$$) the stability of $$+3$$ oxidation state increases down the group and that of $$+5$$ oxidation state decreases down the group, due to inert pair effect.
Hence, the correct options are $$A$$, $$B$$ and $$C$$
Hence, the correct option is $$A$$.
Explanation:
In $$NC{l_5}$$, there is absence of $$d-orbitals$$. $$PC{l_5}$$ has $$d-orbitals$$, so it exists.
Due to inert pair effect $$P{b^{2 + }}$$ is more stable than $$P{b^{4 + }}$$. So it doesn’t form tetravalent compounds.
In carbonate ion the bonds are equal due to the resonance.
The configuration of $$O_2^ + (8 + 8 - 1 = 15) = \sigma 1{s^2},{\sigma ^*}1{s^2},\sigma 2{s^2},{\sigma ^*}2{s^2}$$ $$\sigma 2p_z^2,\pi 2p_x^2 \approx 2p_y^2, {{\pi 2p}^*_{x}}^1 \approx \pi {{2p}^*_{y}}^0$$
The configuration of $$NO (7 + 8 = 15) = \sigma 1{s^2},{\sigma ^*}1{s^2},\sigma 2{s^2},{\sigma ^*}2{s^2}$$ $$\sigma 2p_z^2,\pi 2p_x^2 \approx 2p_y^2, {{\pi 2p}^*_{x}}^1 \approx \pi {{2p}^*_{y}}^0$$
Both have one unpaired electron so they are paramagnetic.
Final answer: The statements B and C are incorrect. So the correct answer is options $$B$$ and $$C.$$
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