Class 12th

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New answer posted

5 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution 

New answer posted

5 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution 

 

New answer posted

5 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

Glucose and fructose are functional group isomers of each other

 

New answer posted

5 months ago

0 Follower 6 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution 

New answer posted

5 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

5 months ago

0 Follower 8 Views

A
alok kumar singh

Contributor-Level 10

[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

New answer posted

5 months ago

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A
alok kumar singh

Contributor-Level 10

[V (CO)6]

EAN = 23 – 0 + 2 * 6

= 23 + 12

= 35 ≠ 36

⇒ so it does not obey EAN rule.

New answer posted

5 months ago

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alok kumar singh

Contributor-Level 10

I2 + conc. HNO3 → HIO3 + NO2

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5 months ago

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alok kumar singh

Contributor-Level 10

H2S2O6

→ Dithionic acid

⇒ There is no S–O–S bond in it.

New answer posted

5 months ago

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A
alok kumar singh

Contributor-Level 10

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

so H2Te will be most acidic among given options.

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