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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 201. |
The most abundant transition metal belong toA. 3d seriesB. 4d seriesC. 5d seriesD. 6d series |
| Answer» Correct Answer - A | |
| 202. |
The number of electrons exchanged when `KMnO_(4)` react with `H_(2)O_(2)` is equal to that ofA. s electrons in Ca ( at. no. : 20)B. p electrons in neon (at. no. : 10)C. d electrons in chromium ( at. No. : 24)D. f electron in lanthanum ( at. no. : 57) |
| Answer» Correct Answer - C | |
| 203. |
Which of the d-block elements are not referred to as the transition elements ?A. Group 9B. Group 12C. Group 10D. Group 13 |
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Answer» Correct Answer - B Transition elements are those d-block elements which have incompletely filled (n-1) d subshells either in their atomic state or in their common ionic states. On the basis of incompletely filled 3d subshell in case of scandium atom in its ground state `(3d^(1) 4s^(2))`, it is regarded as a transition element. On the other hand, Zinc atom (Group. 12) has completely filled 3d subshell `(3d^(10))` in its atomic state as well as in its + 2 state hence it is not regarded as a transition element. Silver atom has completely field (n-1) d subshell in the ground state of its atom `(4d^(10) 5s^(1))`, but it is a transition element because it shows oxidation state of `+ 1(4d^(10))`, `+ 2(4d^(9))` and `+3(4d^(8))`. |
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| 204. |
With O highest possible oxidation state of Mn isA. `+7`B. `+4`C. `+5`D. `+3` |
| Answer» Correct Answer - A | |
| 205. |
The ground state electronic configuration of platinum isA. `[Xe] 4f^(14) 5d^(9) 6s^(1)`B. `[Kr] 4d^(10) 5s^(0)`C. `[Xe] 4f^(14) 5d^(8) 6s^(2)`D. `[Kr] 4d^(4) 5s^(1)` |
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Answer» Correct Answer - A The expected configuratiuon of Pt is `[Xe]4f^(14 5d^(8) 6s^(2)` but its actual confinguration is `[Xe]4f^(14) 5d^(9) 6s^(1)` . Second configuration corresponds to Pd, third one corresponds to Ag and fourth one corresponds to Nb. |
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| 206. |
Which of the following can be reduce easilyA. `V (CO)_(6)`B. `Mo(CO)_(6)`C. `Ni^(+2)`D. `Mn^(+2)` |
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Answer» Correct Answer - A `V (CO)_(6)` easily reduces to `[V (CO)_(6) ]^(+)` |
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| 207. |
Assertion : `HgCl_(2)` and `SnCl_(2)` exist together in an aqueous solution. Reason : `SnCl_(2)` is a strong reducing agent.A. If both the assertion and reason are ture but the reason is a true explanation of the assertion.B. If both the assertion and reason are true but the reason is not the correct explanation of the assertionC. If the assertion is true but reason is falseD. If assertion is false but reason is true. |
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Answer» Correct Answer - D `SnCl_(2)` will reduce `HgCl_(2)` to `HgCl_(2)` and finally to `Hg`. |
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| 208. |
Why is `Cr^(2+)` reducing and `Mn^(3+)` oxidising when both have `d^4` configuration? |
| Answer» `Cr^(2+)` is reducing as its configuration changes from `d^(4)` to `d^(3)`, the latterhaving a half-filled `t_(2g)` level (see Unit 9) . On the other hand, the changefrom `Mn^(3+)` to `Mn^(2+)` results in the half-filled `(d^(5))` configuration which hasextra stability. | |
| 209. |
Name a member of the lanthanoid series which is well known to exhibit `+4` oxidation state. |
| Answer» Correct Answer - Cerium (Z = 58) | |
| 210. |
Which of the following ions will exhibit colour in aqueous solution ?A. `Sc^(3+) (Z = 21)`B. `La^(3+) (Z = 57)`C. `Ti^(3+) (Z = 22)`D. `Lu^(3+) (Z = 71)` |
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Answer» Correct Answer - C Most of the compounds of transition metals are coloured in the solid or in solution states. The color of transition metal ions arises form the excitation of electrons form the d-orbitals of lower energy to the d-orbitals of higher energy. The energy required for d-d electron ecitations is available in the visible range. It is for this reason that transition metal ions have the property to absorb certain radiations form the visible region and exhibit the complementary color. The transition metals ions which have completely empty d-orbitals `(d^(0))` or which have completely filled d-orbitals `(d^(10))` are colorless as the excitation is not possible form one d-orbital to another. Similar inner transition metal ions colored on account of f-f transition which is absent for `f^(0)` or `f^(14)` configuration. `La^(3+) (Z = 57) implies [Xe] 4f^(0) 5d^(0) 6s^(0)` `Ti^(3+) (Z = 22) implies [Ar] 3d^(1) 4s^(0)` `Lu^(3+) (Z = 71) implies [Xe] 4f^(14) 5d^(0) 6s^(0)` `Sc^(3+) (Z = 21) implies [Ar] 3d^(0) 4s^(0)` Thus, only `Ti^(3+)` with one unpaired electron in the subshell exhibits colour. |
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| 211. |
Which of the following oxidation states is the most common among the lanthanoids ?A. 4B. 2C. 5D. 3 |
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Answer» Correct Answer - D The lanthanoids are characterizad by the uniform `(+3)` oxidfation state shown by all the metals. They typically form compounds which are ionic and trivalent. The electronic structures of the ions are `Ce^(3+) f^(1), Pr^(3+) f^(2), Nd^(3+) f^(3), … Lu^(3+) f^(14)` . |
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| 212. |
which of the following electronts is present as the impurity to the maximum extent in the pig iron?A. PhosphorusB. ManganeseC. CarbanD. Silicon |
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Answer» Correct Answer - C Pig iron contains about `4%` carbon. `P`, `Mn` and `Si` are in less percentage. |
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| 213. |
Which one of the elements with the following outer orbital configuration may exhibit the larger number of oxidation states ?A. `3d^(5) 4s^(1)`B. `3d^(5) 4s^(2)`C. `3d^(2) 4s^(2)`D. `3d^(3) 4s^(2)` |
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Answer» Correct Answer - B Variable oxidation states of transition metals arise form fact that the energy difference between (n - 1) d subshell and ns subshell is relatively small and hence electrons form both the subshell can participate in bonding. Mn with maximum no. of unpaired (n - 1) d electrons and maximum no. of ns electrons exhibits the largest numbver of oxibation states: `-1, 0, +1, +2, +3, +4, +5, +6` and `+7`. |
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| 214. |
Which one of the elements with the following outer orbital configuration may exhibit the larger number of oxidation states ?A. `3d^(2), 4s^(2)`B. `3 d^(3), 4 s^(2)`C. `3 d^(5), 4 s^(1)`D. `3 d^(5), 4 s^(2)` |
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Answer» Correct Answer - D It is `Mn` and we know that `Mn` exhibits `+7` oxidation state. |
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| 215. |
Cuprous compounds such as CuCI, CuCN and CuSCN are the only salts stable in water due toA. high hydration energy of `Cu^(+)` ionsB. their inherent tendency not to disproportionateC. diamagnetic natureD. insolubility in water |
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Answer» Correct Answer - D Cuprous compounds are those in which copper is monovalent. Most of the cuprous compounds are colourless and dimagnetic as 3d- subhell is completely filled. However, `Cu_(2)O` and `Cu_(2)S` are and black respectively. Cuprous compounds are generally insoluble in water. The soluble compounds are unstable in aqueous solutions, since they disproportionate to `Cu^(2+)` and Cu. `2Cu^(+) rarr Cu^(2+) + Cu` |
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| 216. |
Transition elements are frequently used as catalyst because:A. or variable oxidationk stateB. of high ionic chargeC. large surface area of reactantsD. of their specific nature |
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Answer» Correct Answer - A Because of unparied which d-electrons, transition metals have variable varlency which enables them to form unstable intermediate compounds. These intermediates give reaction path of lower activation enrergy and, therefore increase the rate of the reaction. |
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| 217. |
Which of the following statements is not correct ?A. `La(OH)_(3)` is less BASIC than `Lu(OH)_(3)`B. In lanthanide series ionic of `Ln^(2+)` ion decreasesC. La is actully an element of transition series rather lanthanide series.D. Atomic radius of Zr and Hf are same because of lanthanide contraction |
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Answer» Correct Answer - A The basic strength of `Ln(OH)_(3)` decreases as the ionic radius decreases form Ce to Lu. Thus `Ce(OH)_(3)` is the most basic, and `Lu(OH)_(3)` , is the least basic. |
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| 218. |
Among `TiF_(6)^(2-), CoF_(6)^(3-), Cu_(2)C1_(2)` and `NiC1_(4)^(2-)` (At. No. `Ti = 22, Co = 27, Cu = 29, Ni = 28)`, the colourless species are -A. `CoF_(6)^(3-)` and `NiCl_(4)^(2-)`B. `TiF_(6)^(2-)` and `CoF_(6)^(3-)`C. `Cu_(2) Cl_(2)` and `NiCl_(4)^(2-)`D. `TiF_(6)^(2-)`, and `Cu_(2) Cl_(2)` |
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Answer» Correct Answer - D `Ti_(6)^(2-)` and `Cu_(2)Cl_(2)` due to absence of unpaired electrons. |
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| 219. |
Which one of the following forms a colourless solution in aqueous medium ? (Atomic number : `Sc =21, Ti = 22, V = 23`, and `Cr = 24`)A. `V^(3+)`B. `Cr^(3+)`C. `Ti^(3+)`D. `Sc^(3+)` |
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Answer» Correct Answer - D Colour is usually due to d-d trnsition : `V (3d^(3) 4s^(2)) implies V^(3+) (3d^(2))` : coloured `Cr (3d^(5) 4s^(1)) implies Cr^(3+) (3d^(3))` : colored `Ti (3d^(2) 4s^(2)) implies Ti^(3+) (3d^(1))` : coloured `Sc (3d^(1) 4s^(2)) implies Sc^(3+) (3d^(0))` : colourless. |
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| 220. |
In which of the following paris both the ions are coloured in aqueous solution? (Atomic number, `Sc - 21, Ti = 22, Ni = 28, Cu = 29, Co = 27)`A. `Ni^(2+)` and `Cu^(+)`B. `Sc^(3+)` and `Ti^(3+)`C. `Sc^(3+)` and `Co^(2+)`D. `Ni^(2+)` and `Ti^(3+)` |
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Answer» Correct Answer - D `Ni^(2+)` and `Ti^(3+)` ions coloured in aqueous solution because they contain unpaired electrons. |
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| 221. |
Cigarette smokers absorb significant levels of ----- from tobacco smoke.A. CdB. ZnC. HgD. Cu |
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Answer» Correct Answer - A Cadmium is a toxic element that is present in our foodstuffs and is normally ingested at levels that are close to the maximum safe leve. The kidney is the orange most susceptible to cadmium, about 200 ppm cause severe damage. Among trace essential elements, Zn is second only to iron in importance. Around teh world a lack of Zn is the most common soil deficiency. Beans, cirtus fruit, coffee, and rice are the crops most susceptible to Zn deficiency. Zn is an essential element for animal. Over 200 Zn enzymes have been identified in living oranisms and their roles determined. Zn enzymes that perform almost every possible type of enzyme function are known, but the most common function is hydrolysis. Hg is hazardous because of its relatively high vapor pressure. The Hg vapor is absorbed in the lungs, dissolves in the blood, and is then carried to the brain, where irreversible damage to the control nevous system results. The metal is also slightly soluble in water, again a result of its very weak metallic bonding. Inorganic compounds of Hg are usually less of a problem because they are not very soluble. However, organomercury compounds pose the greatest danger, as these compounds (such as the methylmercury cation, `HgCH_(3)^(+)`) are readily absorbed and are retaiuned by the body much more strongly than the simple Hg compounds. Chemically, Zn and Cd are rather simple and mercury is somewhat distinct. The compounds of these elements are characterised by the `d^(10)` configuration in which [with the excepetion of `Hg_(2)^(2+)` ion, which formally involves Hg `(+1)`] they almost exclusively involve `M(+2)` . |
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| 222. |
Which of the following statement are correct about `Cr^(2+)` (Z = 24) and `Mn^(3+)` (Z = 25) ? (i) `Cr^(2+)` is a reducing agent (ii) `Mn^(3+)` is an oxidizing agent (iii) Both `Cr^(2+)` and `Mn^(3+)` exhibit `d^(4)` configuration (iv) When `Cr^(2+)` is used as a reducing agent, the chromium ion attains `d^(5)` electronic configurationA. (i), (ii), (iii), (iv)B. (ii), (iii), (iv)C. (i), (ii), (iii)D. (ii), (iii) |
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Answer» Correct Answer - C `Cr(3d^(5) 4s^(1)) rarr Cr^(2+) (3d^(4))` `Mn(3d^(5) 4s^(2)) rarr Mn^(3+) (3d^(4))` Thus both have `d^(4)` configuration. `Cr^(2+)` is a reducing agent (i.e. it reduces the other and gets oxiudized) becasuse it gets oxidized to more stable `Cr^(3+)` . During this transformation th econfiguration of chromium changes form `d^(4)` to `d^(3)`, the latter having a half-filled `t_(2g)` level (strong field as well as weak field ligands) is relatively more stable. `Mn^(3+)` is an oxidizing agent (i.e. oxidizes the other and gets redused) because it gets reduced to more stable `Mn^(2+)`. The change form `Mn^(3+)` to `Mn^(2+)` results in the exactly half-filled `(d^(6))` configuration which has extra stability. Moreover, `d^(5)` configuration leads to half-filled `t_(2g)` level and half-filled `e_(g)` level in case of weake field ligands such as `H_(2)O`. |
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| 223. |
`Cu^(+)` ion is not stable in aqueous solution because because of dispropotionation reaction. `E^(@)` value of disproportionation of `Cu^(+)` is `[E_(Cu^(2+)//Cu^(+))^(@)=+ 0.15 V, E_(Cu^(2+)//Cu)^(@)=0.34 V]`A. second ionisation entyhalpy of copper is less than the first ionisation enthalpyB. large value of second ionisation enthalpy of copper is compensated by much more negative hydration energy of `Cu_((aq))^(2+)`C. hydration energy of `Cu_((aq))^(2+)` is much more negative than that of `Cu_((aq))^(2+)`D. many copper (I) compounds are unstable in aqueous solution and undergo disproportionation rection. |
| Answer» Correct Answer - B | |
| 224. |
Which is least stable in aqueous mediumA. `Fe^(2+)`B. `CO^(+2)`C. `Ni^(+2)`D. `Mn^(+2)` |
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Answer» Correct Answer - A `Fe^(2+)` quickly oxidizes to `Fe^(3+)` in aqueous medium. |
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| 225. |
Conectrated aqueous sodium hydroxide can be a separated mixture ofA. `Al^(3+)` and `Sn^(2+)`B. `Al^(3+)` and `Fe^(3+)`C. `Al^(3+)` and `Zn^(2+)`D. `Zn^(2+)` and `Pb^(2+)` |
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Answer» Correct Answer - B `NaOH + Al^(=3) rarr underset(("Soluble in NaOH"))underset("Sod.metaaluminate")(NaAIO_(2))` `NaOH +Fe^(+3) rarr` No reaction (Insoluble in NaOH) |
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| 226. |
Which of the following is correct for d-block metals ?A. They form exclusively covalent compoundsB. They exclusively form coordination compounds but no simple compoundsC. They may form either ionic or covalent compounds depending on the condition.D. They form only ionic compounds. |
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Answer» Correct Answer - C The ionization enthalpies of d-block metals are greater than those of s-block metals but are lower than those of p-block elements. This implies that the d-block elements are less electropositive than s-block elements but are more electropositive than p-block elements but are more electropositive than p-block elements. Consequently, they may for either ionic or covalent bonds depending on the condition. Generally, the compounds formed in lower valence states are ionic but those in higher valence states are convalent. The first row (3d) elements have many more ionic compounds (on account of relatively low ionization enthalpies) than elements in the second (4d) and third (5d) rows. |
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| 227. |
Which of the following oxoacide of phosphorus is a reducing agent and a monobasic acid as well?A. `H_(3) P_(2) O_(3)`B. `HPO_(3)`C. `H_(3)PO_(3)`D. `H_(3) PO_(2)` |
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Answer» Correct Answer - D `H_(3) PO_(2)` is monobasic and is reducing due to `P - H` bond. |
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| 228. |
The noble charactoer of platinum and gold is favoured byA. high entalpies of sublimation, high ionization enthalpies and high enthapies of solvationB. high enthalpies of sublimation, high ionization enthalpies and low enthalpies of solvationC. high enthalp[ies of sublimation, low ionization enthalpies and low enthalpies of solvationD. low enthalpies of sublimation, high ionization enthalpies and low enthalpies of solvation |
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Answer» Correct Answer - B The d-block metals are less reactive than s-block metals. However, many of the metals are sufficiently electropositive to react with mineral acids, liberating `H_(2)`. A few have low standard electrode potentials and remain unreactive or noble. Except in the ususual circumstances, metals act only as reducing agents. Generally, the reactivity of the d-block metals as reducing agents tends to decreases as we go across the periodic table form left to right. The trend in thier reactivity can be related to their electrode potentoials. Group 3 metals including lanthanides and actinides are strong reducing agents. The metals of Groups 4-7 are moderately reactive as are Fe, Ru, Os, Co and Ni of Groups 8 to 10. The remaining metals of Groups 8-10, Rh, Ir, Pt and Pd, as well as Ag and Au have low reactivity. Hence, they are called noble metals. Noble character is favoured by (i) high enthalpies of sublimation, high ionization ehthalpies and low enthalpies of solvation. The high melting points lead to high enthalpies of solvation. The smaller atoms result in high ionization enthalpies. This tendency top noble character is most pronounced for the platinum metals (Ru, Rh, Pd, Os, Ir, Pr) and gold. |
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| 229. |
The correct condiguration of f-block elements isA. `(n-2)f^(1-14)(n-1)d^(0-1)ns^2`B. `(n-1)f^(1-14)(n-1)d^(0-1)ns^2`C. `(n-3)f^(1-14)(n-2)d^(0-1)(n-1)s^2`D. `(n-2)f^(0-1)(n-1)d^(0-1)ns^2` |
| Answer» Correct Answer - A | |
| 230. |
Which of the following statement is correct ?A. Because of lanthanide contraction, the separation of second-row d-block elements form one another is easierB. Because of lanthanide contraction, the radii of the third-row d-block elements are almost the same as those of the second-row elementsC. Because of lanthanide contraction, the radii of the third-row elements are almost the same as those of the first-row elements.D. The second-row elements have smaller radii than the corresponding third-row elements. |
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Answer» Correct Answer - B The group trends in atomic radii of the d-block elements are parallel to those observed in s- and p- block elements. As we go down the group, there is an increas in atomic size up to the second series of d-block elements. This is expected as the electrons enter the new shell of electrons in the second d-block series. But the atomic radii of the third (5d) series are virtually the same as those of the corresponding members of the second (4d) series. This phenomenon is associated with the intervention of the 4f orbitals. The filling 4f before 5d subshell results in a regular decrease in atomic radii called lanthaboid which essentally compensates for the expected increase in atomic size with increase atomic number. The net result of the lanthanoid contraction is that the second and the third d series exihibit similar radii (e.g., Zr 160 pm, Hf 159 pm) and have very similar physical andchemical properties. |
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| 231. |
Among the following the compound that is both paramagnetic and coloured isA. `K_(2)Cr_(2)O_(7)`B. `(NH_(4))_(2) [TiCl_(6)]`C. `VOSO_(4)`D. `K_(3) [Cu(CN)_(4)]` |
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Answer» Correct Answer - C (a) `Cr_(2) O_(7)^(2-) Cr^(+6)` is `d^(0)` (b) `2 xx 1 + x + 6 xx - 1 = 0` `x = 4` `Ti^(4+)` is `d^(0)` (c ) `x - 2 - 2 = 0` `x = 4` `V^(4+)` is `d^(1)` (d) `3 xx 1 + x + 4 xx - 1 = 0` `x = + 1` `Cu^(+)` is `d^(10)` |
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| 232. |
Which of the d-block elements has the lowest electronegativity ?A. CoB. NiC. CuD. Zn |
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Answer» Correct Answer - D d-block elements have fairly low values of electronegativity. In 3d-series it generally increases form Sc to Cu with a fall at Zn. However, this increase in electronegativity is much slower because the additional electron is being added to an inner shell which provides relatively good shielding to the outermost ns electrons form the nucleus. Consequently, the elements (Co, Ni and Cu) in the second half of the 3d- series have similar electronegativity. The increasing electronegativity form Sc to Cu impiles that the elements becomne slightly less metallic and this is reflected in the decreasing negative `E^(Theta)(M^(2+)//M` and `M_(3+)//M^(2+))` values. |
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| 233. |
Which of the following d-block elements has the bigger covalent radius ?A. CoB. NiC. CuD. Zn |
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Answer» Correct Answer - B Co (116 pm) gt Ni (115 pm) lt Cu (117 pm) lt Zn (125 pm) As we move form alkali metals to alkaline earth metals and form alkain earth metals to the d-block metals, the atomic radii decrease steeply but within d-block elements this rate of decrease is less due to increase in the number of inner shielding electrons. However, the general trend of decreasing size is reversed towerds the ends of the series due to an increase in inter-electronic repulsion a after the addition of sufficient number of electrons in the (n-1) d orbitals leading to the gradual increase in size. Other examples Rh(125 pm) lt pd(128) lt Ag(134) lt Cd(141) Ir(126 pm) lt Pt(129) lt Au(134) lt Hg(144) |
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| 234. |
What will the structure of `CrO_(5)` in presence of pyridine ?A. ButterflyB. Square pyramidalC. Pentagonal pyramidalD. Cannot be predicted |
| Answer» Correct Answer - C | |
| 235. |
A compound of a metal ion `M^(X+)(z=24)` has a spin only magnetic moment of `sqrt(15)B.M.` . The number of unpaired electrons in the compound areA. 2B. 4C. 5D. 3 |
| Answer» Correct Answer - D | |
| 236. |
Which of the following lantanide ion is paramagnetic ?A. `Ce^(4+)`B. `Yb^(2+)`C. `Lu^(3+)`D. `Eu^(2+)` |
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Answer» Correct Answer - D `Ceto [Xe]4f^1 5d^1 6s^2, Ce^(4+)to [Xe]` `Yb to [Xe]4f^14 6s^2, Yb^(2+)to[Xe] 4f^14` `Luto[Xe]4f^14 5d^1 6s^2, Lu^(3+)to[Xe]4f^14` `Euto[Xe]4f^7 6s^2, Eu^(2+)to [Xe]4f^7` |
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| 237. |
Compound that is both paramagnetic and coloured is:A. `K_2Cr_2O_7`B. `(NH_4)_2[TiCl_6]`C. `VOSO_4`D. `K_3[Cu(CN)_4]` |
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Answer» Correct Answer - C `K_2Cr_2O_7" contains " Cr^(6+)(3d^0)`, which is diamagnetic but coloured due to charge transfer spectra. `(NH_4)_2[TiCl_6]" contains "V^(4+)(3d^1)`, which is paramagnetic and coloured. `K_3[Cu(CN)_4]" contain " Cu^(+)(3d^10)`, which is diamagnetic and colourless. |
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| 238. |
`CuSO_4` is paramagnetic while `ZnSO_4` is diamagnetic becauseA. `Cu^(2+)` ion has `3d^9` configuration while `Zn^(2+)` ion has `3d^(10)` configurationB. `Cu^(2+)` ion has `3d^5` configuration while `Zn^(2+)` ion has `3d^6` configurationC. `Cu^(2+)` has half filled orbitals while `Zn^(2+)` has fully filled orbitalsD. `CuSO_4` is blue in colour while `ZnCO_4` is white. |
| Answer» Correct Answer - A | |
| 239. |
Which of the following statement is correct regarding 3d-series of elements ? (1) Sc has the lowest while Zn has the highest first ionization enthlpy. (2) Cu has the highest while Sc has the lowest second ionization enthalpy (3) Zn has the highest third ionization enthalpy (4) Fe has lower third ionization enthalpy relative to MnA. (i), (ii), (iii)B. (ii), (iii)C. (i), (ii), (iii), (iv)D. (i), (iv) |
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Answer» Correct Answer - C In the case of d-block metals, the variation of ionization enthalpy across the periods and down the groups parallel quite closely the trend in atomic size. As we move across a period, the effective nuclear charge experienced by ns electrons goes on increasing causing the shells to shrink in size and thus making it difficult to removed the electrons. Thus along a period the ionization enthalpy generally increases. The second and third ionization enthalpies follow the same pattern, except for the second ionization enthalpies of Cr and Cu which are comparatively higher due to extra ionization enthalpies of Mn and Zn are comparatively higher due to extra stability of `3d^(5)` and `3d^(10)` configuration. Similarly the third ionization enthalpies of Mn and Zn are comparatively higher due to extra stability of `3d^(5)` and `3d^(10)` configuration. As the decrease in the size of atoms of the d-block metals is less than that of the main group elements along a period, the ionization enthalpies to the increase along the series only slightly as compared to the main group elements. For example, in 3d-series the elements Ti, V and Cr have similar first ionization enthalpies. Similarly the elements Fe, Co, Ni and Cu have similar first ionization enthalpies. Since (n-1) d and ns electrons do not differ much in energy, the difference in the successive ionization enthalpies is relatively small. As we move down a group form the elements of first d series to those of the second d series, there is a decrease in the ionization enthalpy on account of increasing size. But it again increases when we move further down the group from second to the third d-series. This trend is consistent with relatively small size of the atoms of elements of the third d series due to the insertion of the lanthanides which causes the third row d-block elements to have grater than expected effective nuclear charge. |
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| 240. |
Which of the following statements is correct about stability of the complexes of lanthanoids?A. Stability of complexes increases as the size of lanthanoid decreases.B. Stability of complexes decreases as the size of lanthanoid decreases.C. Lanthanoids do not form complexes.D. All the complezes of lanthanoids have same stability. |
| Answer» Correct Answer - A | |
| 241. |
Which of the following lanthanoids has highest tendency to form complexes ?A. `Ce^(+3_`B. `Pm^(+2)`C. `Lu^(+3)`D. `Eu^(+2)` |
| Answer» Correct Answer - C | |
| 242. |
Stability of the complex may depend onA. Ionisation energyB. Hydration energyC. Sublimation energyD. All of these |
| Answer» Correct Answer - D | |
| 243. |
The correct regarding `CuCl_(5)^(-3)` compound isA. Hybridisation is `sp^(3)d`B. Axial bond length is larger than equitorial bond lengthC. Equatorial bond length is longer than axial bond lengthD. Both (1) & (3) |
| Answer» Correct Answer - D | |
| 244. |
The species which is paramagneticA. `Cr^(+)`B. `Zn^(2+)`C. `Cu^(+)`D. `MnO_(2)O^(-)` |
| Answer» Correct Answer - A | |
| 245. |
The species which convert `Cu^(2+)` to `Cu^(+)`A. `I^(-)`B. `HCHO //OH^(-)`C. `CN^(-)`D. All of these |
| Answer» Correct Answer - D | |
| 246. |
Amongst the following ions, which is considered as most stable in `M^(2+)` state ?A. `Ti^(2+)//Ti(-1.63V)`B. `V^(2+)//V(-1.11V)`C. `Cr^(2+)//Cr(-0.90V) `D. `Mn^(2+)//Mn(1.18V)` |
| Answer» Correct Answer - A | |
| 247. |
Electrode potential of `Mn^(2+)//M` for Ni is abnormal because ofA. High `IE_(1)+IE_(2)`B. High hydration energyC. `DeltaH_("atomisation")`D. Electronic configuration of `Ni^(2+)` |
| Answer» Correct Answer - B | |
| 248. |
The hybridisation of Cu in `(NH_(4))_(2)[CuCl_(4)] and Cs_(2)[CuCl_(4)]` isA. `dsp^(2)` in bothB. `dsp^(2) and sp^(3)` respectivelyC. `sp^(3) and dsp^(2)` respectivelyD. `sp^(3)` in both |
| Answer» Correct Answer - B | |
| 249. |
When `K_(2)MnO_(4)` is added in solution of `NH_(4)Cl` thenA. Green colour will appearB. Yellow colour will appearC. Pink colour will appearD. Colour will appear |
| Answer» Correct Answer - B | |
| 250. |
`KMnO_(4)` dissollution in concentration `H_(2)SO_(4)` results in explosion due toA. Formation of MnO which explodeB. Formation of `Mn_(2)O_(7)` which explodeC. Formation of `MnO_(2)` which explodeD. Formation of `MnSO_(4)` which explode |
| Answer» Correct Answer - B | |