The d- and f- Block Elements Set-1

$4.90\ \mathrm{BM}$

$5.92\ \mathrm{BM}$

$3.87\ \mathrm{BM}$

$0.00\ \mathrm{BM}

Square planar $[Ni(CN)_4]^{2-}$ is paramagnetic with two unpaired electrons; octahedral $[Ni(H_2O)_6]^{2+}$ is diamagnetic.

Both $[Ni(CN)_4]^{2-}$ and $[Ni(H_2O)_6]^{2+}$ are paramagnetic with two unpaired electrons.

Both complexes are diamagnetic because $d^8$ always pairs in octahedral and square planar fields.

$[Ni(CN)_4]^{2-}$ is diamagnetic due to strong-field CN$^-$ causing pairing (square-planar $d^8$), while $[Ni(H_2O)_6]^{2+}$ is high-spin and paramagnetic with two unpaired electrons.

$[Fe(CN)_6]^{4-}$ has larger crystal-field stabilization energy and is diamagnetic (low-spin $d^6$), whereas $[Fe(H_2O)_6]^{2+}$ is high-spin ($d^6$) and paramagnetic.

$[Fe(H_2O)_6]^{2+}$ has larger CFSE and is diamagnetic; $[Fe(CN)_6]^{4-}$ is paramagnetic.

Both complexes are high-spin and paramagnetic with similar CFSE.

$[Fe(H_2O)_6]^{2+}$ is low-spin and diamagnetic while $[Fe(CN)_6]^{4-}$ is high-spin and paramagnetic.

$d^3$ and low-spin $d^6$

high-spin $d^5$ and $d^2$

high-spin $d^4$ and $d^9$

$d^8$ and $d^{10}$

Au$^+$ is more electronegative than Cu$^+$ so it cannot be oxidized further, preventing disproportionation.

Relativistic contraction and stabilization of gold's 6s/5d orbitals enhance ligand covalency and stabilize Au(I) linear complexes; Cu$^+$ lacks such relativistic stabilization and has less ligand-stabilization relative to its redox tendency, so Cu$^+$ disproportionates.

Copper has a partially filled d-shell in Cu$^+$ which makes it intrinsically unstable and leads to disproportionation.

Gold readily forms Au(II) which prevents disproportionation of Au(I) in solution.

$1.73\ \mu_B$

$5.92\ \mu_B$

$3.87\ \mu_B$

$4.90\ \mu_B$

$360\ \mathrm{nm}$

$800\ \mathrm{nm}$

$2250\ \mathrm{nm}$

$1800\ \mathrm{nm}$

Cr is in oxidation state $+3$, electronic configuration $d^3$, with 3 unpaired electrons

Cr is in oxidation state $+2$, electronic configuration $d^4$, with 4 unpaired electrons

Cr is in oxidation state $+3$, electronic configuration $d^4$, with 2 unpaired electrons

Cr is in oxidation state $+2$, electronic configuration $d^3$, with 3 unpaired electrons

$Mn^{2+}$ has higher charge density than $Fe^{3+}$ and therefore produces a more acidic solution

Both ions produce comparable acidity in water since they are isoelectronic

$Fe^{3+}$, having higher charge and smaller radius, has larger (more exothermic) hydration enthalpy and polarises coordinated water more strongly, so $Fe^{3+}$ hydrolyses more and gives a more acidic solution

$Fe^{3+}$ hydrolyses less than $Mn^{2+}$ because its contracted $3d$ electrons reduce its polarising power

$Ce^{4+}$ is stabilised because its $4f$ electrons are strongly core-like and therefore unavailable for bonding

By attaining an empty $4f$ shell ($4f^0$) equivalent to the noble-gas core $[Xe]$, and because the $+4$ charge gives very large lattice and hydration energies together with increased covalent/polarising interactions, $Ce^{4+}$ is comparatively stabilised

$Ce^{4+}$ is stable mainly because Ce has the smallest ionic radius among the series, maximising crystal-field stabilisation

$Ce^{4+}$ is stabilised because a half-filled $4f^7$ subshell is achieved, giving extra stability