Ncert Solutions Chemistry Class 12th

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alok kumar singh

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Kindly go through the solution

 

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alok kumar singh

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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

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alok kumar singh

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A jump is seen after 2nd Ip so Ve = 2

hence configuration would be ns2

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Photodiode in reverse bias mode is used as intensity measuring device.

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Rainbow is formed due to internal reflection and dispersion.

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In hypermetropia image formed by eye lens is beyond retina and is focussed to retina by a convex lens.

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alok kumar singh

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Kindly go through the solution

 

New answer posted

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alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

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