- ✓$p-$ block, group $=\, 13$
- B$p-$ block, group $=\, 14$
- C$d-$ block, group $=\, 9$
- D$s-$ block, group $=\, 2$
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Reason : The only property that determines its aromatic behaviour is its planar structure.
$4M + 8CN^-+ 2H_2O + O_2 \rightarrow 4[M(CN)_2]^-+ 4OH^-$
$2[M(CN)_2]^-+ Zn \rightarrow [Zn(CN)_4]^{2-} + 2M$
| Molecule | $P-X$ (axial) bond length | $P-X$ (Equitorial) bond length |
| $PF_5$ | $a$ | $b$ |
| $PF_4CH_3$ | $c$ | $d$ |
| $PF_3 (CH_3)_2$ | $e$ | $f$ |
| $PCl_5$ | $g$ | $h$ |
According to given information choose the incorrect order of bond length
$\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}_{2}$ $\xrightarrow[{{\text{H}}_{2}}{{\text{O}}_{2}},\bar{O}\text{H}]{\text{B}{{\text{H}}_{3}},\text{THF}}(P)$ $\xrightarrow[\text{C}{{\text{H}}_{2}}\text{C}{{\text{l}}_{2}}]{\text{ Pyridinium Chloro Chromate }(\text{PCC})}$ $(Q)$
$\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}_{2}$ $\xrightarrow[NaB{{H}_{4}}\,,\,H{{O}^{\Theta }}]{Hg\,{{(OAc)}_{2}}\,,\,{{H}_{2}}O}(R)$ $\xrightarrow[\text{C}{{\text{H}}_{2}}\text{C}{{\text{l}}_{2}}]{\text{ Pyridinium Chloro Chromate }(\text{PCC})}$ $(S)$