The calculated heat of formation $(\Delta \mathrm{H_f})$ for $\mathrm{NaCl_2}$ and $\mathrm{CaF}$ is
(a)negative for both $\mathrm{NaCl_2}$ and $\mathrm{CaF}$
(b)negative for $\mathrm{NaCl_2}$ but positive for $\mathrm{CaF}$
(c)positive for $\mathrm{NaCl_2}$ but negative for $\mathrm{CaF}$
(d)positive for both $\mathrm{NaCl_2}$ and $\mathrm{CaF}$
Answer
SELF-PRACTICE — the source book printed no answer.
Nothing is invented here, so this question has no answer on record.
Explanation
62) a — $s$ and $p$ both form $\pi$-bonding with the complex, and the $\pi$-bonding capacity of sulphur is greater than phosphorus due to the smaller size of the d-orbital of sulphur. Hence, in the presence of sulphur trans to the phosphorus donor atom, the phosphorus-metal bond will be weak, hence they do not lie trans to each other in the complex. As nitrogen does not involve in $\pi$-bonding with the complex, when a nitrogen atom is trans to phosphorus, phosphorus becomes able to form an efficient $\pi$-bond with the metal, hence becomes stable; that's why P and N are trans to each other.63) c64) a — Terminal NO $=1672\,\mathrm{cm^{-1}}$; Bridging NO $=1505\,\mathrm{cm^{-1}}$65) d — N: $1s\,2s\,2p$; $m_l = +1, 0, -1$; $S = \tfrac{1}{2}+\tfrac{1}{2}+\tfrac{1}{2} = \tfrac{3}{2}$; $L = 1+0-1 = 0$.66) a — Trans effect $\mathrm{NO_2} > \mathrm{Cl} > \mathrm{NH_3}$67) c — Total isomers = geometrical isomers + $\tfrac{1}{2}$(optical isomers). Geometrical isomers = 2 isomers (trans form + cis form). Optical isomers = 2 isomers (d-cis-form + l-cis form). Total isomers = 368) a — There is spin change during the electron transition. Therefore, the transition is spin disallowed.69) a —
Base hydrolysis of $[\mathrm{CoCl(NH_3)_5}]^{2+}$ depends on the concentration of both $[\mathrm{CoCl(NH_3)_5}]^{2+}$ and base. In this reaction $\mathrm{OH^-}$ abstracts a proton from coordinated $\mathrm{NH_3}$. The base hydrolysis of $[\mathrm{Co(CN)}]^{3-}$ is independent of base because it has no ionizable proton.70) c — $[\mathrm{Ne}]\,2p^{1}\,3p^{1}$; $2S+1=3$; $2S=2$; $S=1$. For $^3D$: $L=2$, $J=|L+S|,\dots,|L-S| = |2+1|,\dots,|2-1| = 3,2,1$. Levels for $^3D$ are $^3D_3$, $^3D_2$, $^3D_1$.71) a — $[\mathrm{MnF_6}]^{3-}$, $\mathrm{Mn^{3+}} \rightarrow 3d^4 \rightarrow t_{2g}^{3}e_g^{1}$ (HS). $\mu = \sqrt{4(4+2)} = \sqrt{24} = 4.9$. $S=2 \rightarrow 2S+1=5$; $L=2 \rightarrow D$. Ground state term symbol $^{3}D$.72) a — It is the first inorganic optically active compound discovered by Werner. Number of Co–O bonds = 12. Number of Co–N bonds = 12.73) a — $[\mathrm{CrCl_6}]^{3-}$, $\mathrm{Cr^{3+}} \rightarrow 3d^3$. $\mu_s = \sqrt{n(n+2)} = \sqrt{3(3+2)} = 3.87$. $\mu_{L+S} = \sqrt{n(n+2)+l(l+1)}$ BM $= \sqrt{3(3+2)+3(3+1)} = \sqrt{15+12} = \sqrt{27} = 5.20$ BM.74) a — The bonding in phosphine ligands, like that of carbonyls, has two components. The primary component is $\sigma$ donation of the phosphine lone pair to an empty orbital on the metal. The second component is back donation from a filled metal orbital to an empty orbital on the phosphine ligand. This empty phosphorous orbital has been described as being either a d-orbital or an antibonding sigma orbital ($\sigma^{*}$). So, in transition metal phosphine (M–PR$_3$) complexes, the back bonding involves donation of electrons from $M(t_{2g}) \rightarrow \mathrm{PR_3}(\sigma^{*})$.75) b — For allowed transition: (a) Spin multiplicity should be the same, or $\Delta S = 0$. Option c and d (incorrect) as $\Delta S \neq 0$. (b) $\Delta L = \pm1$, for allowed transition $^{3}S \rightarrow {}^{3}P$: $\Delta S = 3-3=0$; $\Delta L= 1-0= 1$.76) d — $ns^1\,np^1$; $L=1 \equiv P$; $S=1$ hence, $2S+1=3$. For $J$ value, if the orbital is less than half filled then $J=|L-S|=|1-1|=0$. Hence, $^{3}P_0$.77) a — $\mathrm{Fe(II)}$ protoporphyrin $\rightarrow \pi \rightarrow \pi^{*}$ transition. $[\mathrm{Mn(H_2O)_6}]\mathrm{Cl_2} \rightarrow \mathrm{Mn^{2+}} \rightarrow d^5$ (high spin), spin forbidden (d-d) transition. $\Delta S=1$. Hence, not allowed. $[\mathrm{Co(H_2O)_6}]\mathrm{Cl_2} \rightarrow \mathrm{Co^{2+}} \rightarrow d^7$ (high spin), (d-d), $\Delta S= 0$, hence allowed.78) c — $\mathrm{K_4[Cr(CN)_6]}= \mathrm{Cr^{2+}} \rightarrow d^4$ (low spin). For a compound to show Jahn-Teller distortion, the $e_g$ or $t_{2g}$ set should be electronically degenerate. (1) $\mathrm{Cr^{2+}} \rightarrow d^4$ (low spin), electronically degenerate, hence shows Jahn-Teller distortion. (2) $\mathrm{K_4[Fe(CN)_6]} = \mathrm{Fe^{2+}}$ (low spin) $= d^6$ (low spin), electronically non-degenerate, hence no Jahn-Teller distortion. (3) $\mathrm{K_3[Co(CN)_6]} = \mathrm{Co^{3+}} \rightarrow d^6$ (low spin), same as above, hence no Jahn-Teller distortion. (4) $\mathrm{K_4[Mn(CN)_6]} = \mathrm{Mn^{2+}}$ (low spin), electronically degenerate, hence the complex will show Jahn-Teller distortion.79) a — $[\mathrm{Fe(Phen)_2(NCS)_2}]$, $\mathrm{Fe^{2+}} = d^6$ complex. At high temperature, high spin, and at low temperature, low spin behaviour. At 250 K: $\mathrm{CFSE} = 1.6\Delta_0 +1.2\Delta_0 =0.44$, $\mu = \sqrt{n(n+2)} = \sqrt{4(4+2)} = 4.90$. At 150 K: $\mathrm{CFSE} = -0.4 \times 6\,\Delta_0 = -2.4\Delta_0$, $\mu =0$.80) a — Taking the valence shell configuration $\mathrm{Be}: 2s^1\,3s^1$; $S=\tfrac12+\tfrac12 =1$. Multiplicity $=2S+1= 2\times1+1=3$. $L= 0+0= 0$, an $S$ term. $J= (L+S) = (0+1) = 1$. Hence, the term is $^{3}S_1$.81) a — $[\mathrm{CrO_4}]^{2-} < [\mathrm{MnO_4}]^{2-} < [\mathrm{FeO_4}]^{2-}$. All have +6 oxidation state, but due to the smaller size on going from Cr to Fe, Fe has very high charge density. Hence, it has a very high tendency to accept an electron. Hence, it is the strongest oxidising agent.82) c — Number of unpaired electrons: (A) $[\mathrm{CoF_6}]^{3-} \rightarrow d^6$ (high spin) $= t_{2g}^{4}e_g^{2}$ (B) $[\mathrm{IrCl_6}]^{3-} \rightarrow d^6$ (low spin) $= t_{2g}^{6}e_g^{0}$ (C) $[\mathrm{Fe(H_2O)_6}]^{2+} \rightarrow d^6$ (high spin) $= t_{2g}^{4}e_g^{2}$. Since, A and C have the same number of unpaired electron, hence they have the same magnetic moment. As A and C are 3d-block metal with weak ligand, hence they are high spin while Ir, being a 5d-metal, is low spin. Because 5d has greater splitting power than 3d.83) a — $[\mathrm{Mn(H_2O)_6}]^{2+} = d^6$ (HS). As all the levels are electronically non-degenerate, hence no Jahn-Teller distortion. Therefore, all the Mn–O bond lengths will be equal.84) a — $[\mathrm{Fe(phen)_3}]^{2+} \rightarrow [\mathrm{Fe(phen)_3}]^{3+} + e^-$ (Red) (Blue). In presence of an oxidising agent like $\mathrm{K_2Cr_2O_7}$, $[\mathrm{Fe(phen)_3}]^{2+}$ changes its colour from red to blue, hence they are used in redox titration.85) a — The stability (formation constant for complexation) of a cryptate complex depends upon (1) size of cavity (2) size of metal cation. Hence, $\mathrm{K^+}$ ion will form the most stable complex with cryptand-222.86) c — $E_2 > E_1$, hence energy required for transfer of electron from oxygen to Re falls in the UV region. Hence, $[\mathrm{ReO_4}]^{-}$ colourless, while in $[\mathrm{MnO_4}]^{-}$, due to less energy difference, it falls in the visible region. Hence, $[\mathrm{MnO_4}]^{-}$ coloured. Also as both Mn and Re have +7 oxidation state, i.e. they have no d-electron, hence no d-d transition, and colour arises due to LMCT.87) a —
The reaction proceeds through a dissociative pathway, and the rate is dependent only on the concentration of the substrate. As the substrate does not have an acidic hydrogen, hence it does not undergo an $S_N1CB$ mechanism.
88) a — Explanation89) c — Tanabe-Sugano diagrams are useful in interpretation of spectra of both high spin and low spin complexes of $d^{2}$-$d^{8}$ metal cations. In a Tanabe-Sugano diagram, the energy of excited state (expressed as $E/B$) is plotted against ligand field strength (expressed as $\Delta_0/B$). Zero energy is taken for the lowest term and also two terms of the same symmetry never cross each other and they bend apart from each other due to repulsion.90) b — $[\mathrm{CrF_6}]^{3-} = d^3$. Orgel diagram for $d^3$: ground state $= {}^4F$, first excited state $= {}^4P$. $^4A_{2g} \rightarrow {}^4T_{2g}$ ($\Delta_0$): $\nu_1 = 14900\,\mathrm{cm^{-1}}$; $\nu_2 = 22700\,\mathrm{cm^{-1}}$; $\nu_3 = 34400\,\mathrm{cm^{-1}}$. Since $^4T_{1g}(P)$ and $^4T_{1g}(F)$ have no fixed energy, therefore they will not provide an accurate value of $\Delta_0$. Thus, the energy difference between $^4A_{2g} \rightarrow {}^4T_{2g}$, $14900\,\mathrm{cm^{-1}}$, will correspond to $\Delta_0$.91) b — For $1s^{2}2s^{2}2p^{6}3p^{1}$: $2S+1 = 2\times\tfrac12+1 = 2$; $J=|l+s|,\dots,|l-s| = |1+\tfrac12|,\dots,|1-\tfrac12| = \tfrac32, \tfrac12$. Term symbols $^2P_{3/2}$, $^2P_{1/2}$.92) b — $^3F \rightarrow {}^3D$. Since, for allowed transition (atomic), $\Delta S=0$, $\Delta L=0,\pm1$. In option a and c, spin multiplicity is not the same, therefore incorrect. In option d, $\Delta L=2$, therefore incorrect.93) d — $[\mathrm{Ni^{II}L_6}]^{n+\,\text{or}\,n-}$ shows absorption bands at $8500$, $15400$, and $26000\,\mathrm{cm^{-1}}$. $[\mathrm{Ni^{II}L'_6}]^{n+\,\text{or}\,n-}$ at $10750$, $17500$ and $28200\,\mathrm{cm^{-1}}$. As for the first complex, absorption bands are at low energy, i.e. it has weak splitting, therefore it has a weak ligand, and for the second complex, high energy absorption bands correspond to a strong ligand. Thus, $L$ is weak and $L'$ is strong ligand.94) b — Number of microstates in $^3F$ is calculated as $(2S+1)(2L+1)$. For $F$, $L=3$. Hence, $(3)(2\times3+1) =21$.95) b — Chelate effect is predominately due to entropy change. $\Delta H$ nearly same but entropy change is more.96) b — $d^6$: $L= \sum m_L = 5-3 = 2 = D$. $S=2$, $2S+1=5$. Hence, lowest energy form $= {}^5D$.97) d — $\mathrm{H_2}$ (excited state): $L=0=\Sigma$; $S=\tfrac12+\tfrac12= 1$, $2S+1=3$; term $= {}^{3}\Sigma_u$. For half filled.98) c — $[\mathrm{Cr(bipyridyl)_3}]^{3+} \rightarrow d^3 = {}^4F$ (ground state term). As phosphorescence is a spin-forbidden transition and also occurs when electron comes from excited to ground state, hence the transition $^4A_{2g} \leftarrow {}^2E_g$ is responsible for the phosphorescence.99) c — For orbital contribution, the set should be unsymmetrically filled. $[\mathrm{Cu(H_2O)_6}]^{2+} \rightarrow t_{2g}^{6}e_g^{3}$ → no contribution. $[\mathrm{Ni(H_2O)_6}]^{2+} \rightarrow t_{2g}^{6}e_g^{2}$ → no contribution. $[\mathrm{Co(H_2O)_6}]^{2+} \rightarrow t_{2g}^{5}e_g^{2}$ → orbital contribution. $[\mathrm{Cr(H_2O)_6}]^{2+} \rightarrow t_{2g}^{3}e_g^{1}$ → no contribution.