Chemistry NCERT Exemplar Solutions Class 12th Chapter Nine

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Payal Gupta

Contributor-Level 10

During the electrolysis of dilute H2SO4

2H2SO4 (1)electrolysis2HSO4+2H+

H2S2O8+2H2Ohydrolysis2H2SO4+H2O2 (A)

2H++2eH2

In the solid form of H2O2 dihedral angle is equal to 90.2°.

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alok kumar singh

Contributor-Level 10

In reducing action, H2O2 changes to O2 because it will oxides but option (A) is in acidic medium, hence answer will be (C).

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alok kumar singh

Contributor-Level 10

              K2 [Cu (CN)4]

              Oxidation number of Cu is +1

              Cu+ = [Ar]3d10 ® Diamagnetic

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Payal Gupta

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Tritium is radioactive and it decays into He3 during emission of β-radiation

1T32He3 + -1e0

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Payal Gupta

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Stress = Y * strain

T1A=Y* (l1l)l (i)

T2A=Y* (l2l)l (ii)

 T1T2=l1ll2l

l=T1l2T2l2T1T2=T2l1T1l2T2T1

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Vishal Baghel

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Kindly consider the following figure

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alok kumar singh

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Cr2+ and Fe2+ both have 4 unpaired electrons.

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Payal Gupta

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All metal carbonyls have synergic bonds.

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Payal Gupta

Contributor-Level 10

  [Ni (CN)4]2

CN is strong field ligand

Ni2+=4s03d8

Here;  [Ni (CN)4]2 is square planar and diamagnetic.

[Ni (CO)4]

Ni = 4s23d8Co is strong field ligand.

Here ;  [Ni (CO)4] is tetrahedral and diamagnetic

[Ni (CN)4]2 has 3d8 configuration while  [Ni (CO)4] has 3d10 configuration.

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