Ncert Solutions Chemistry Class 12th

Hence, electronic configuration of CO²⁺ is $\left[Ar\right]3d^{7}4s^0$ . In complex ${\left[Co{\left(CN\right)}_6\right]}^{4-},$ given ligand CN^- is $\therefore$ strong hence, after pairing in d-subshell, total number of unpaired electron =

spin magnetic moment = $\sqrt[]{1\left(1+2\right)}=\sqrt[]{3}=1.73BM$ $=1.73BM\approx 2.0BM$  

Both statement are correct for the glass body heating, but reason is not correct explanation during heating process of glass, constituents unit rupture of glass body and gives the edge smoothness.

All structure can produce -CHO functional group which gives +ve test of Tollen's reagent except structure (II), because if forms ketonic group after tautomerisation

$X\underset{H_{2}S{O}_4}{\overset{Conc}{\to }}$ Brown fumes + Brown ring test with FeSO₄/H₂SO₄

$x\underset{HCl}{\overset{H_{2}S}{\to }}Y\left(ppt\right)\underset{HN{O}_3}{\overset{Conc}{\to }}dissolved\underset{N{H}_{4}OH}{\overset{}{\to }}$ Deep blue colour solution

In cation analysis of Cu⁺ ions, precipitate formed is CuS on treating with H₂S and HCl which dissolved in HNO₃ and produced blue colour complex solution ${\left[Cu{\left(N{H}_3\right)}_4\right]}^{++}$ in NH₄OH. Brown fumes and brown ring performance given by nitrate sample. Hence salt is Cu (NO₃)₂.

In $Br{O}_4^{-},Br$ has +7 O.S, therefore it only acts as a oxidizing reagent, does not undergo disproportionation reaction

In fcc structure of diamond four C present in fcc lattice and other four C present in tetrahedral voids where 50% of tetrahedral voids are occupied. Hence number of carbon atoms present per unit cell of diamond is 8.

$NiC{l}_2+NaCN\underset{O.A.}{\overset{strong}{\to }}{\left[Ni{\left(CN\right)}_6\right]}^{2-}$

Complex has Ni⁴⁺ and strong ligand, hence following are the metal ion electronic configuration

Change of unpaired electron = 2

Wt of Cl^- in 100 ml = 1.8 * 10^-3 gm

Mol. of Cl^- in 100 ml = $\frac{1.8*10^{-3}}{35.5}=0.0507*10^{-3}mole$  

$\therefore \left[C{l}^{-}\right]=\frac{0.0507*10^{-3}*1000}{100}=5.07*10^{-4}M$   

i.e. 0.507 milli mole in one lit required in one hr.

Coagulation value = (millimole/lit) required in one hr = 0.507

= 1 (the nearest integer)

Here, $\lambda =\frac{h}{\sqrt[]{2*m*eV}}=\frac{6.63*10^{-34}}{\sqrt[]{2*1.6*10^{-19}*40000*9.1*10^{-31}}}m$  

$\therefore \lambda =0.614*10^{-11}m$  

$\lambda =6.14*10^{-12}m$ [the nearest integer = 6]