Question: From NCERT | NEET || [Difficult level:Easy] ||
The correct order of decreasing second ionization enthalpy of Ti(22), V(23), Cr (24) and Mn(25) is
1. Cr > Mn > V > Ti
2. V > Mn > Cr > Ti
3. Mn > Cr > Ti > V
4. Ti > V > Cr > Mn_
: 1. Cr > Mn > V > Ti..
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NCERT REFERENCE
NCERT REFERENCE
It has
been observed that half filled and fully filled
degenerate set of orbitals acquire extra stability
due to their symmetry. (XI Page 63)
New pattern is followed beginning with
scandium (Sc). The 3d orbital, being lower in
energy than the 4p orbital, is filled first.
Consequently, in the next ten elements,
scandium (Sc), titanium (Ti), vanadium (V),
chromium (Cr), manganese (Mn), iron (Fe),
cobalt (Co), nickel (Ni), copper (Cu) and zinc
(Zn), the five 3d orbitals are progressively
occupied. We may be puzzled by the fact that
chromium and copper have five and ten
electrons in 3d orbitals rather than four and
nine as their position would have indicated with
two-electrons in the 4s orbital. The reason is
that fully filled orbitals and half-filled orbitals
have extra stability (that is, lower energy). Thus
p3, p6, d5, d10,f 7, f14 etc. configurations, which
are either half-filled or fully filled, are more
stable. Chromium and copper therefore adopt
the d5 and d10 configuration(XI Pg 64).
The valence electronic configurations of
Cr and Cu, therefore, are 3d5 4s1 and 3d10 4s1
respectively and not 3d4 4s2 and 3d9 4s2. It has
been found that there is extra stability
associated with these electronic configurations.[XI pg 64]____________________
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