The 'spin only' magnetic moment value of MO42- is BM. (Where M is a metal having least metallic radii. among Sc, Ti, V, Cr, Mn and Zn). (Given atomic number: Sc = 21, Ti = 22, V = 23, Cr = 24, Mn = 25 and Zn = 30)
0
Metallic radii (pm): Sc 164, Ti 147, V 135, Cr 129, Mn 137, Zn 137; the least is Cr, so M = Cr.
In CrO42-: x + 4(−2) = −2, so x = +6
Cr6+ = [Ar] 3d0: no unpaired electron (n = 0)
∴ μ = n(n + 2) = 0 BM
Answer: 0
The equilibrium Cr2O72- 2CrO42- is shifted to the right in :
Given below are two statements : In the light of the above statements, choose the correct answer from the options given below :
The number of elements from the following that do not belong to lanthanoids is Eu, Cm, Er, Tb, Yb and Lu
The difference in the 'spin-only' magnetic moment values of KMnO4 and the manganese product formed during titration of KMnO4 against oxalic acid in acidic medium is BM. (nearest integer)
Among CrO, Cr2O3 and CrO3, the sum of spin-only magnetic moment values of basic and amphoteric oxides is 10-2 BM (nearest integer). (Given atomic number of Cr is 24 )
Arrange the following elements in the increasing order of number of unpaired electrons in it. Choose the correct answer from the options given below :
Among VO2+, MnO4- and Cr2O72-, the spin-only magnetic moment value of the species with least oxidising ability is BM (Nearest integer). (Given atomic number V = 23, Mn = 25, Cr = 24)
The metal that shows highest and maximum number of oxidation state is :
The spin-only magnetic moment value of the ion among Ti2+, V2+, Co3+ and Cr2+, that acts as strong oxidising agent in aqueous solution is BM (Near integer). (Given atomic numbers : Ti : 22, V : 23, Cr : 24, Co : 27)
The number of ions from the following that have the ability to liberate hydrogen from a dilute acid is . Ti2+, Cr2+ and V2+