Explanation
Cr = 3d54s1; After removing 4s1 electron, the next electron is to be removed from relatively more stable half filled 3d5 subshell which will require more energy for its ionization.
MnO2 + 4KOH + O2 → 2K2MnO4 + 2H2O
Purple green
Na3VO4 contains vanadium in + 5 oxidation state which has vacant d- orbitals and is thus colourless.
Y is the first member of 4d-transition series. Its size is less than the size of lanthanoids. In case of lanthanoids the size decreases from La3+ to Lu3+ Thus, the correct order of ionic radii is Y3+ < Lu3+ < Eu3+ < La3+.
Iodide of Millon’s base (a dark brown ppt.) obtained by the action of alkaline K2HgI4 is NH2HgO.HgI.
Vanadyl ion is VO2+ where V is in + 4 oxidation state.
Size increases in the order Fe3+ < Fe2+ < Mn2+ < Ti2+ as the nuclear charge decreases.
Mn2O7 +H2O → 2HMnO4. It is permanganic acid which is violet in colour.
Sc3+(3do), Ti4+(3do) are diamagnetic while Pd2+(4d8) and Cu2+(3d9) are paramagnetic.
‘925 fine silver’ means 925 parts of pure Ag in 1000 parts of an alloy. Therefore, in percentage it will be 92.5% Ag and 7.5% Cu.
Mn2+ in MnSO4. 4H2O has d5 configuration (five unpaired electrons); Cu2+ in CuSO4. 5H2O has d9 configuration (one unpaired electron); 2Fe+ in FeSO4.6H2O has d6 configuration) (four unpaired electrons) and Ni2+ in NiSO4. 6H2O has d8 configuration (two unpaired electrons). Thus CuSO4. 5 H2O has lowest degree of paramagnetism.
K2MnO4 is formed when MnO2 is fused with KOH in air
2 MnO2 + 4 KOH + O2 → 2 K2MnO2 + 2 H2O
K2Cr2O7 contains Cr6+ (3d0) ion which is diamagnetic and coloured due to charge transfer. (NH4)2 [TiCl6] contains.Ti4+ (3d10) which is diamagnetic and colourless. VOSO4 contains V4+ (3d1) which is paramagnetic and coloured. K3[Cu(CN)4] contains Cu+(3d10) which is diamagnetic and colourless.
The existence of Fe2+ and NO+ in nitroprusside ion, [Fe(CN)5NO]2- can be established by measuring the magnetic moment of the solid compound which should correspond to (Fe2+ = 3d6) four unpaired electrons.
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