Advertisements
Advertisements
Question
How would you account for the following?
Transition metals exhibit variable oxidation states.
Advertisements
Solution
(i) The variable oxidation states of transition elements are due to the participation of ns and (n−1)d-electrons in bonding. Lower oxidation state is exhibited when ns-electrons take part in bonding. Higher oxidation states are exhibited when (n − 1) d-electrons take part in bonding.
APPEARS IN
RELATED QUESTIONS
Calculate magnetic moment of `Fe_((aq))^(2+) ion (Z=26).`
Out of Mn3+ and Cr3+, which is more paramagnetic and why ?
(Atomic nos. : Mn = 25, Cr = 24)
Calculate the number of unpaired electrons in the following gaseous ions:
Mn3+, Cr3+, V3+ and Ti3+. Which one of these is the most stable in an aqueous solution?
Write down the number of 3d electrons in the following ion:
Cu2+
Indicate how would you expect the five 3d orbitals to be occupied for this hydrated ion (octahedral).
Complete and balance the following chemical equations
`Fe^(2+) + MnO_4^(-) + H^+ ->`
Which among the following transition metal has the lowest melting point?
Out of \[\ce{Cu2Cl2}\] and \[\ce{CuCl2}\], which is more stable and why?
A solution of \[\ce{KMnO4}\] on reduction yields either a colourless solution or a brown precipitate or a green solution depending on pH of the solution. What different stages of the reduction do these represent and how are they carried out?
The product of oxidation of I– with \[\ce{MnO^{-}4}\] in alkaline medium is:-
Write the ionic equation for reaction of KI with acidified KMnO4.
