The reason for greater range of oxidation states in actinoids is attributed to
The radioactive nature of actinoids
Actinoid contraction
5f, 6d and 7s levels having comparable energies
5f, 6d and 7s levels having comparable energies
The correct increasing order of ionic radii of the following Ce3+, La3+, Pm3+ and Yb3+ is
Yb3+ < Pm3+< Ce3+<La3+
Ce3+< Yb3+<Pm3+<La3+
Yb3+< Pm3+< La3+<Ce3+
Pm3+ < La3+< Ce3+ < Yb3+
The shape of gaseous SnCl2 is
Tetrahedral
Linear
Angular
T-shape
C.
Angular
Total no. of valence electron for Sn = 4
Total no. of valence electron for Cl = 7
Thus total number of valence electron for SnCl2 = 4 + 7 x 2 = 18
Bond pair = 2
Number of electrons used to complete the octets and duet = 8 x 2 = 16
No. of lone pair = 1
SnCl2 has sp2 hybridisation with one lone pair of electrons. Thus has angular shape.
[CuCl4]2- exists while [CuI4]2- does not exist, because
I- is stronger reductant than Cl-
I- is weaker reductant than Cl-
I- is stronger oxidant than Cl-
None of the above
Actinides exhibit larger number of oxidation state than that of corresponding lanthanides. The reason behind this aspect is
lesser energy difference between 5f and 6d-orbitals than between 4f and 5d-orbitals
larger atomic size of actinides than the lanthanides
more energy difference between 5f and 6d orbitals than between 4f and 5d-orbitals
greater reactive nature of the actinides than the lanthanides
Acidified K2Cr2O7 solution turns green when Na2SO3 is added to it. This is due to the formation of
Cr
Cr2(SO3)3
CrSO4
Cr2(SO4)3
Point out the incorrect reaction from the following.
2Na2CrO4 + H+ Na2Cr2O7 + 2Na+ + H2O
4MnO2 + 4KOH + O2 4KMnO4 + 2H2O
2Mn + 5C2 + 16H+ 2Mn2+ + 10CO2 + 8H2O
Mn + 8H+ + 5Fe2+ 5Fe3+ + Mn2+ + 4H2O
On addition of small amount of KMnO4 to conc. H2SO4, a green oily compound is obtained which is highly explosive in nature. Identify the compound from the following
Mn2O7
MnO2
MnSO4
Mn2O3