- A$K_2Cr_2O_7$
- B$(NH_4)_2[TiCl_6]$
- ✓$VOSO_4$
- D$K_3[Cu(CN)_4]$
$Cr ^{-6}=[ Ar ] 3 d ^0$
In $\left( NH _4\right)_2\left[ TiCl _6\right]$, the oxidation state of $Ti$ is $+4$.
$Ti ^{+4}=[ Ar ] 3 d ^0$
In $VOSO _4$, the oxidation state of $V$ is $+4$.
$V ^{+4}=[ Ar ] 3 d ^1$
In $K _3\left[ Cu ( CN )_4\right]$, the oxidation state of $Cu$ is $+1$.
$Cu ^{+1}=[ Ar ] 3 d ^{10}$
In $VOSO _4, V ^{+4}$ configuration is $d ^1$, so it has one unpaired d-electron so it is paramagnetic and colored. All other compounds don't have any unpaired $d$ electrons, so they are diamagnetic in nature.
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$Q$ is
($1$) Which of the following options has correct combination considering $List-I$ and $List-II$?
$(1) (III), (S), (R)$ $(2) (IV), (Q), (R)$ $(3) (III), (T), (U)$ $(4) (IV), (Q), (U)$
($2$) Which of the following options has correct combination considering $List-I$ and $List-II$?
$(1) (I), (Q), (T), (U)$ $(2) (II), (P), (S), (U)$ $(3) (II), (P), (S), (T)$ $(4) (I), (S), (Q), (R)$
Give the answer quetion ($1$) and ($2$)
| Column $I$ | Column $II$ |
| $(A) \,XX '$ | $(i)$ $T-$shape |
| $(B)\,XX'_3$ | $(ii)$ Pentagonal bipyramidal |
| $(C)\,XX '_5$ | $(iii)$ Linear |
| $(D)\,XX '_7$ | $(iv)$ Square pyramidal |
| $(v)$ Tetrahedral |