The dand fblock elements - Result Question 34
####34. The common oxidation states of Ti are [1994]
(a) $+2,+3$
(b) $+3,+4$
(c) $-3,-4$
(d) $+2,+3,+4$
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Solution:
- (b) $Ti(22)=[Ar] 3 d^{2} 4 s^{2}$
minimum oxidation state is +2 and maximum oxidation state is +4 .
$Ti^{4+}=[Ar]$
Hence, it is most stable state of Ti.
$Ti^{3+}=[Ar] 3 d^{1}$
It acts as a good reducing agent as it can readily oxidized to $Ti^{4+}$ ion.
$Ti^{2+}=[Ar] 3 d^{2}$
It is the most unstable form.
Thus, +3 and +4 are the common oxidation states. $\frac{2}{\text{ 이 }}$
The minimum oxidation state of transition metals is equal to the number of electrons in $4 s$ shell and the maximum oxidation state is equal to the sum of the $4 s$ and $3 d$ electrons.
