Class 12th hindi Arihant book up board

Kindly go through the solution 

Glucose and fructose are functional group isomers of each other

Kindly go through the solution 

[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