MCQ
${p_x}$ orbital can accommodate
- A$4$ electrons
- B$6 $ electrons
- C$2$ electrons with parallel spins
- ✓$2 $ electrons with opposite spins
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$I ^{-}+ ClO _3^{-}+ H _2 SO _4 \longrightarrow Cl ^{-}+ HSO _4^{-}+ I _2$
The correct statement(s) in the balanced equation is/are:
$(A)$ Stoichiometric coefficient of $HSO _4^{-}$is 6 .
$(B)$ lodide is oxidized.
$(C)$ Sulphur is reduced.
$(D)$ $H _2 O$ is one of the products.

| List$-I$ | List$-II$ | ||
| $(P)$ | $[Cr(NH_3)_4Cl_2]Cl$ | $(1)$ | Paramagnetic and exhibits ionisation isomerism |
| $(Q)$ | $[Ti(H_2O)_5Cl](NO_3)_2$ | $(2)$ | Diamagnetic and exhibits cis-trans isomerism |
| $(R)$ | $[Pt(en)(NH_3)Cl]NO_3$ | $(3)$ | Paramagnetic and exhibits cis-trans isomerism |
| $(S)$ | $[Co(NH_3)_4(NO_3)_2]NO_3$ | $(4)$ | Diamagnetic and exhibits ionisation isomerism |