Class 12th

[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

The quantisation of electric charge, q = ne, applies to electric charge only, even though charge cannot exist without mass.

One simple rule to think here is that electric charge is a scalar quantity with magnitude. It has positive and negative signs, depending on the direction it is forced to move in an electric field. Mass is always positive. So when you add mass, it never cancels out or becomes zero. Also do consider that a charge can never exist when there is no mass. In calculations, you must remember not to take in the mass but just the charge itself.