- A$Si$
- ✓$Pb$
- C$Ge$
- D$Sn$
This is most stated in Groups 13,14 and 15
A possible given explanation is that due to presence of f-subshell electrons are to diffused, due to the shape of the forbitals, to effectively shield the s electrons from the pull of the nucleus.
A good example would be the elements of Group $13 .$ Aluminium in +1 state is unknown, and the stability of +1 oxidation state increases as we go down the group.
Thallium is most stable in +1 oxidation state, and compounds of Thallium in +3 oxidation state are known to be very strong oxidising agents, and have a tendency to reduce quickly to +1 state.
Hence option B is correct.
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$(A)$ $\mathrm{Ni}(\mathrm{CO})_{4}$
$(B)$ $\left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{2}$
$(C)$ $\mathrm{Na}_{2}\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]$
$(D)$ $\mathrm{PdCl}_{2}\left(\mathrm{PPh}_{3}\right)_{2}$

$(a)$ Effective nuclear charge for nitrogen is $3.90$
$(b)$ $IP$ of $Ne$ is more than $Na^+$
$(c)$ Order of electron negativity $sp > sp^2 > sp^3$
$(d)$ Order of acidic character $NH_3 < PH_3 < AsH_3$