Identify correct statements from below:
A. The chromate ion is square planar.
B. Dichromates are generally prepared from chromates.
C. The green manganate ion is diamagnetic.
D. Dark green coloured K₂ MnO₄ disproportionates in a neutral or acidic medium to give permanganate.
E. With increasing oxidation number of transition metal, ionic character of the oxides decreases.
Choose the correct answer from the options given below:

Solution

Chemistry ma’am · 4 steps

Check each statement against the oxidation state and the d-electron count, then match the true set.

1. Shape and preparation. In CrO₄²⁻, chromium is +6 and bonds to four oxygens arranged as a tetrahedron, not a square plane, so A is false. Dichromate is made by acidifying chromate, 2CrO₄²⁻ + 2H⁺ → Cr₂O₇²⁻ + H₂O, so B is true.
2. Manganate chemistry. In MnO₄²⁻, manganese is +6, which leaves one unpaired d electron (d¹). So manganate is paramagnetic and C is false. In neutral or acidic solution it disproportionates to permanganate and MnO₂, so D is true.

   3MnO₄²⁻ + 4H⁺ → 2MnO₄⁻ + MnO₂ + 2H₂O

3. Oxide character. As the oxidation number of the metal rises, it pulls electron density back from oxygen more strongly, so the bonds become more covalent. For example, MnO is ionic and basic, while Mn₂O₇ is a covalent, acidic oxide. So E is true.

   Statement	Verdict	Reason
   A	False	CrO₄²⁻ is tetrahedral
   B	True	acidifying chromate gives dichromate
   C	False	Mn(VI) is d¹, so it is paramagnetic
   D	True	disproportionates to MnO₄⁻ and MnO₂
   E	True	higher oxidation state gives more covalent oxides

4. So. The true statements are B, D and E, which is option (a).

The idea to keep. For any oxoanion, find the metal's oxidation state first. It gives you the d-electron count, and so the magnetism, and it tells you how covalent and acidic the oxide is.

Answer from the NTA final answer key. Solution by PrepWiser.