Chemistry

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[CrF₆]^–4

⇒ Cr+2 → 3d4

F– → WFL → No pairing so unpaired e^– = 4

(b)  [MnF₆]^–4

Mn⁺² → 3d⁵

F– → WFL → No pairing unpaired e– = 5

(c)  [Cr (CN)₆]^–4 ⇒ Cr⁺² ⇒ d⁴ CN^– → SFL

→ unpaired e– = 2

(d)  [Mn (CN)₆]–4 ⇒ 3d5,

unpaired e– = 1

[V (CO)₆]

EAN = 23 – 0 + 2 * 6

= 23 + 12

= 35 ≠ 36

⇒ so it does not obey EAN rule.

I₂ + conc. HNO₃ → HIO₃ + NO₂

H₂S₂O₆

→ Dithionic acid

⇒ There is no S–O–S bond in it.

Acidic nature of hydrides increases down the group in p-block

so H₂Te will be most acidic among given options.

A jump is seen after 2nd Ip so Ve^– = 2

hence configuration would be ns²

SO₃ (s) exist as cyclic trimer ⇒ S₃O₉

→ sp3, so all bond π–bonds would be of pπ–dπ so pπ–pπ bond = 0

pπ–dπ bonds = 6

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