Pair of transition metal ions having the same number of unpaired electrons is : [2025]
(1)
Ion | Electronic configuration | Number of unpaired electrons |
3 | ||
3 | ||
2 | ||
5 | ||
4 | ||
1 | ||
5 |
The correct decreasing order of spin only magnetic moment values (B.M.) of , , and ions is : [2025]
(3)
Ion | Electronic configuration | Number of unpaired electrons (n) | Spin only magnetic moment () BM |
0 | 0 | ||
1 | |||
4 | |||
3 |
Thus, the decreasing order of spin-only magnetic moment is:
Match List I with List II. [2025]
List I | List II | ||
(Transition metal ion) | (Spin only magnetic moment (B.M.)) | ||
A. | I. | 3.87 | |
B. | II. | 0.00 | |
C. | III. | 1.73 | |
D. | IV. | 2.84 |
Choose the correct answer from the options given below:
A–III, B–I, C–IV, D–II
A–III, B–I, C–II, D–IV
A–IV, B–II, C–III, D–I
A–II, B–IV, C–I, D–III
(1)
Transition metal ion | Electronic configuration | Number of unpaired electrons (n) | Spin only magnetic moment ()BM |
1 | |||
3 | |||
2 | |||
0 |
The correct option with order of melting points of the pairs (Mn, Fe), (Tc, Ru) and (Re, Os) is : [2025]
Mn < Fe, Tc < Ru and Os < Re
Fe < Mn, Ru < Tc and Re < Os
Mn < Fe, Tc < Ru and Re < Os
Fe < Mn, Ru < Tc and Os < Re
(1)
In any row the melting points of these metals rise to a maximum at except for anomalous values of Mn and Tc and fall regularly as the atomic number increases. The maxima at about the middle of each series indicate that one unpaired electron per orbital is particularly favourable for strong interatomic interaction. In general, greater the number of valence electrons, stronger is the resultant bonding. Because of stable half filled configurations, Mn and Tc have lesser tendency to participate in metallic bonding, hence have lower melting points than expected value.
The element that does not belong to the same period of the remaining elements (modern periodic table) is : [2025]
Platinum
Osmium
Palladium
Iridium
(3)
GP-8 | GP-9 | GP-10 | |
P-5 | Pd | ||
P-6 | Os | Pt |
The number of valence electrons present in the metal among Cr, Co, Fe and Ni which has the lowest enthalpy of atomisation is [2025]
8
9
1
10
(3)
Element | Cr | Fe | Co | Ni |
Ionization enthalpy (kJ/mol) | 397 | 416 | 425 | 430 |
Cr has lowest enthalpy of atomization among given options.
Cr has 6 valence electrons.
The first transition series metal ‘M’ has the highest enthalpy of atomisation in its series. One of its aquated ions () exists in green colour. The nature of the oxide formed by the above ion is : [2025]
neutral
acidic
basic
amphoteric
(3)
In first transition series (3d series) Vanadium has highest enthalpy of atomization. is green in colour. Oxide formed by i.e. is basic in nature.
Among Sc, Mn, Co and Cu, identify the element with highest enthalpy of atomisation. The spin only magnetic moment value of that element in its +2 oxidation state is _____ B.M (in nearest integer). [2025]
(4)
Element | Sc | Mn | Co | Cu |
Enthalpy of atomization (kJ/mol) | 326 | 281 | 425 | 339 |
Co has highest enthalpy of atomization among Sc, Mn, Co and Cu.
, Number of unpaired electrons in = 3.
Spin only magnetic moment ()
Niobium (Nb) and ruthenium (Ru) have and number of electrons in their respective 4d orbitals. The value of is _______. [2025]
(11)
The spin only magnetic moment () value (B.M.) of the compound with strongest oxidising power among , TiO and VO is _______ B.M. (Nearest integer). [2025]
(5)
As reduction potential of is more than that of and , is a stronger oxidising agent than VO and TiO.
, number of unpaired electrons () = 4.
Spin only magnetic moment ()
(nearest integer is 5)