Check statement A: "The chromate ion is square planar."

Chromate ion is $\mathrm{CrO_4^{2-}}$.

• Chromium is surrounded by four oxygen atoms.
• Its geometry is tetrahedral, not square planar.

Therefore, A is false.

Check statement B: "Dichromates are generally prepared from chromates."

Yes. On acidifying chromate solution, dichromate is formed:

$2\,\mathrm{CrO_4^{2-}} + 2\,\mathrm{H^+} \rightleftharpoons \mathrm{Cr_2O_7^{2-}} + \mathrm{H_2O}$

Therefore, B is true.

Check statement C: "The green manganate ion is diamagnetic."

Manganate ion is $\mathrm{MnO_4^{2-}}$.

Oxidation state of Mn:

$x + 4(-2) = -2 \implies x = +6$

So Mn is $\mathrm{Mn^{6+}}$, with electronic configuration $3d^1$? Let us check carefully:

• Mn: $[\mathrm{Ar}]\,3d^5 4s^2$
• $\mathrm{Mn^{6+}}$: remove 2 electrons from $4s$ and 4 from $3d$, giving $[\mathrm{Ar}]\,3d^1$

Since it has one unpaired electron, it is paramagnetic, not diamagnetic.

Therefore, C is false.

Check statement D: "Dark green coloured $\mathrm{K_2MnO_4}$ disproportionates in a neutral or acidic medium to give permanganate."

This is correct. Manganate(VI) disproportionates in neutral or acidic medium:

$3\,\mathrm{MnO_4^{2-}} + 4\,\mathrm{H^+} \rightarrow 2\,\mathrm{MnO_4^-} + \mathrm{MnO_2} + 2\,\mathrm{H_2O}$

Thus permanganate is formed.

Therefore, D is true.

Check statement E: "With increasing oxidation number of transition metal, ionic character of the oxides decreases."

This is a standard trend.

• Lower oxidation state oxides are more ionic.
• Higher oxidation state oxides are more covalent and often acidic.

Example for vanadium/chromium/manganese oxides supports this.

Therefore, E is true.

Collect the correct statements:

• A: False
• B: True
• C: False
• D: True
• E: True

Hence the correct set is:

$\boxed{\text{B, D, E only}}$

Match with options:

Option A = B, D, E only.

So the correct option is:

$\boxed{A}$

Comparison with stored correct answer:

Stored correct answer = A

Our derived answer = A

Hence, they agree.