Electronic configuration of a transition element X in +3 oxidation state is [Ar]3d . What is its atomic number?
- (i)25
- (ii)26
- (iii)27
- (iv)24
Show answer
(ii) 26
Class 12 Chemistry · 71 questions · 69 with answers
Electronic configuration of a transition element X in +3 oxidation state is [Ar]3d . What is its atomic number?
(ii) 26
The electronic configuration of Cu(II) is 3d 9 whereas that of Cu(I) is 3d10. Which of the following is correct?
(i) Cu(II) is more stable
Metallic radii of some transition elements are given below. Which of these elements will have highest density? Element Fe Co Ni Cu Metallic radii/pm 126 125 125 128
(iv) Cu
Generally transition elements form coloured salts due to the presence of unpaired electrons. Which of the following compounds will be coloured in solid state?
(ii) CuF2
On addition of small amount of KMnO4 to concentrated H2SO4, a green oily compound is obtained which is highly explosive in nature. Identify the compound from the following.
(i) Mn2O7
The magnetic nature of elements depends on the presence of unpaired electrons. Identify the configuration of transition element, which shows highest magnetic moment.
(ii) 3d5
Which of the following oxidation state is common for all lanthanoids?
(ii) +3
Which of the following reactions are disproportionation reactions? (a) Cu+ → Cu2+ + Cu – + – (b) 3MnO4 + 4H → 2MnO4 + MnO2 + 2H2O (c) 2KMnO4 → K2MnO4 + MnO2 + O2 – + (d) 2MnO4 + 3Mn2+ + 2H2O → 5MnO2 + 4H
(i) a, b
When KMnO4 solution is added to oxalic acid solution, the decolourisation is slow in the beginning but becomes instantaneous after some time because
(iv) Mn2+ acts as autocatalyst.
There are 14 elements in actinoid series. Which of the following elements does not belong to this series?
(iii) Tm
KMnO4 acts as an oxidising agent in acidic medium. The number of moles of KMnO4 that will be needed to react with one mole of sulphide ions in acidic solution is
(i)
Which of the following is amphoteric oxide? Mn2O7, CrO3, Cr2O3, CrO, V2O5, V2O4
(i) V2O5, Cr2O3
Gadolinium belongs to 4f series. It’s atomic number is 64. Which of the following is the correct electronic configuration of gadolinium?
(i) [Xe] 4f 75d 6s2
Interstitial compounds are formed when small atoms are trapped inside the crystal lattice of metals. Which of the following is not the characteristic property of interstitial compounds?
(iv) They are chemically very reactive.
The magnetic moment is associated with its spin angular momentum and orbital angular momentum. Spin only magnetic moment value of Cr3+ ion is ___________.
(ii) 3.87 B.M.
KMnO4 acts as an oxidising agent in alkaline medium. When alkaline KMnO4 is treated with KI, iodide ion is oxidised to ____________.
(iii) IO3 –
Which of the following statements is not correct?
(i) Copper liberates hydrogen from acids.
When acidified K2Cr2O7 solution is added to Sn2+ salts then Sn2+ changes to
(iii) Sn4+
Highest oxidation state of manganese in fluoride is +4 (MnF4) but highest oxidation state in oxides is +7 (Mn2O7) because ____________.
(iv) in covalent compounds fluorine can form single bond only while oxygen forms double bond.
Although Zirconium belongs to 4d transition series and Hafnium to 5d transition series even then they show similar physical and chemical properties because___________.
(iii) both have similar atomic radius.
Why is HCl not used to make the medium acidic in oxidation reactions of KMnO4 in acidic medium?
(ii) KMnO4 oxidises HCl into Cl2 which is also an oxidising agent.
Generally transition elements and their salts are coloured due to the presence of unpaired electrons in metal ions. Which of the following compounds are coloured?
(i) KMnO4
(ii) Ce (SO4)2
Transition elements show magnetic moment due to spin and orbital motion of electrons. Which of the following metallic ions have almost same spin only magnetic moment?
(i) Co2+
(iv) Cr3+
In the form of dichromate, Cr (VI) is a strong oxidising agent in acidic medium but Mo (VI) in MoO3 and W (VI) in WO3 are not because ___________.
(ii) Mo(VI) and W(VI) are more stable than Cr(VI).
(iii) Higher oxidation states of heavier members of group-6 of transition series are more stable.
Which of the following actinoids show oxidation states upto +7?
(ii) Pu
(iv) Np 0–2 2
General electronic configuration of actionoids is (n–2)f 1–14 (n-1)d ns .Which of the following actinoids have one electron in 6d orbital?
(i) U (Atomic no. 92)
(ii) Np (Atomic no.93)
Which of the following lanthanoids show +2 oxidation state besides the characteristic oxidation state +3 of lanthanoids?
(ii) Eu
(iii) Yb
Which of the following ions show higher spin only magnetic moment value?
(ii) Mn2+
(iii) Fe 2+
Transition elements form binary compounds with halogens. Which of the following elements will form MF3 type compounds?
(i) Cr
(ii) Co
Which of the following will not act as oxidising agents?
(ii) MoO3
(iii) WO3
Although +3 is the characteristic oxidation state for lanthanoids but cerium also shows +4 oxidation state because ___________.
(ii) it has a tendency to attain noble gas configuration
(iii) it has a tendency to attain f 0 configuration
Why does copper not replace hydrogen from acids?
Cu shows positive E value.
Why E values for Mn, Ni and Zn are more negative than expected?
Hint : Negative E values for Mn2+ and Zn2+ are related to stabilities of half filled and fully filled configuration respectively. But for Ni2+ , E value is related to the highest negative enthalpy of hydration.
Why first ionisation enthalpy of Cr is lower than that of Zn ?
Ionisation enthalpy of Cr is lower due to stability of d 5 and the value for Zn is higher because its electron comes out from 4s orbital.
Transition elements show high melting points. Why?
The high melting points of transition metals are attributed to the involvement of greater number of electrons in the interatomic metallic bonding from (n-1) d-orbitals in addition to ns electrons
When Cu2+ ion is treated with KI, a white precipitate is formed. Explain the reaction with the help of chemical equation.
Hint : Cu2+ gets reduced to Cu+ – 2Cu2+ + 4I → Cu2I2 + I2 (white precipitate)
Out of Cu2Cl2 and CuCl2, which is more stable and why?
Hint : CuCl2 is more stable than Cu2Cl2. The stability of Cu2+ (aq.) rather than Cu+(aq.) is due to the much more negative ∆hydH of Cu2+ (aq.) than + Cu (aq.).
When a brown compound of manganese
(A) is treated with HCl it gives a gas
(B) . The gas taken in excess, reacts with NH3 to give an explosive compound
(C) . Identify compounds A, B and C.
Although fluorine is more electronegative than oxygen, but the ability of oxygen to stabilise higher oxidation states exceeds that of fluorine. Why?
Hint : It is due to the ability of oxygen to form multiple bonds to metals.
Although Cr3+ and Co2+ ions have same number of unpaired electrons but the magnetic moment of Cr3+ is 3.87 B.M. and that of Co2+ is 4.87 B.M. Why?
Hint : Due to symmetrical electronic configuration there is no orbital contribution in Cr3+ ion. However appreciable orbital contribution takes place in Co2+ ion.
Ionisation enthalpies of Ce, Pr and Nd are higher than Th, Pa and U. Why?
Hint : It is because in the beginning, when 5f orbitals begin to be occupied, they will penetrate less into the inner core of electrons. The 5f electrons will therefore, be more effectively shielded from the nuclear charge than 4f electrons of the corresponding lanthanoids. Therefore outer electrons are less firmly held and they are available for bonding in the actinoids.
Although Zr belongs to 4d and Hf belongs to 5d transition series but it is quite difficult to separate them. Why?
Hint : Due to lanthanoid contraction, they have almost same size (Zr, 160 pm) and (Hf, 159 pm).
Although +3 oxidation states is the characteristic oxidation state of lanthanoids but cerium shows +4 oxidation state also. Why?
It is because after losing one more electron Ce acquires stable 4f 0 electronic configuration.
Explain why does colour of KMnO4 disappear when oxalic acid is added to its solution in acidic medium.
KMnO4 acts as oxidising agent. It oxidises oxalic acid to CO2 and itself changes to Mn2+ ion which is colourless. 2– – + 5C2O4 + 2MnO4 + 16H → 2Mn2+ + 8H2O + 10CO2 (Coloured) (Colourless) 2– OH– 2–
When orange solution containing Cr2O72– ion is treated with an alkali, a yellow solution is formed and when H+ ions are added to yellow solution, an orange solution is obtained. Explain why does this happen?
Cr2 O7 CrO 4 H+ Dichromate Chromate (Orange) (Yellow)
A solution of KMnO4 on reduction yields either a colourless solution or a brown precipitate or a green solution depending on pH of the solution. What different stages of the reduction do these represent and how are they carried out?
Oxidising behaviour of KMnO4 depends on pH of the solution. In acidic medium (pH < 7) MnO4– + 8H+ + 5e– → Mn2+ + 4H2O (Colourless) In alkaline medium (pH>7) MnO4– + e– → MnO42– (Green) In neutral medium(pH=7) – – MnO4 + 2H2O + 3e– → MnO2 + 4OH (Brown precipitate)
The second and third rows of transition elements resemble each other much more than they resemble the first row. Explain why?
Due to lanthanoid contraction, the atomic radii of the second and third row transition elements is almost same. So they resemble each other much more as compared to first row elements.
E of Cu is + 0.34V while that of Zn is – 0.76V. Explain.
Hint : High ionisation enthalpy to transform Cu(s) to Cu2+ (aq) is not balanced by its hydration enthalpy. However, in case of Zn after removal of electrons from 4s-orbital, stable 3d10 configuration is acquired.
The halides of transition elements become more covalent with increasing oxidation state of the metal. Why?
As the oxidation state increases, size of the ion of transition element decreases. As per Fajan’s rule, as the size of metal ion decreases, covalent character of the bond formed increases.
While filling up of electrons in the atomic orbitals, the 4s orbital is filled before the 3d orbital but reverse happens during the ionisation of the atom. Explain why?
n + l rule : For 3d = n + l = 5 4s = n + l = 4 So electron will enter in 4s orbital. Ionisation enthalpy is responsible for the ionisation of atom. 4s electrons are loosely held by the nucleus. So electrons are removed from 4s orbital prior to 3d.
Reactivity of transition elements decreases almost regularly from Sc to Cu. Explain. 111 d- and f- Block Elements
Hint : It is due to regular increase in ionisation enthalpy. 117 d- and f- Block Elements IV. Matching Type
Match the catalysts given in Column I with the processes given in Column II. Column I (Catalyst) Column II (Process) (i) Ni in the presence
(a) Zieglar Natta catalyst of hydrogen (ii) Cu2Cl2
(c) Vegetable oil to ghee (iv) Finely divided iron
(b) Contact process (iii) V2O5
(d) Sandmeyer reaction (v) TiCl4 + Al (CH3)3 (e) Haber’s Process (f) Decomposition of KClO3
Match the compounds/elements given in Column I with uses given in Column II. Column I (Compound/element) Column II (Use) (i) Lanthanoid oxide
(a) Production of iron alloy (ii) Lanthanoid
(b) Television screen (iii) Misch metal
(c) Petroleum cracking (iv) Magnesium based alloy is
(d) Lanthanoid metal + iron constituent of (v) Mixed oxides of (e) Bullets lanthanoids are employed (f) In X-ray screen
Match the properties given in Column I with the metals given in Column II. Column I (Property) Column II (Metal) (i) An element which can show
(a) Mn +8 oxidation state (ii) 3d block element that can show
(c) Os (iii) 3d block element with highest
(b) Cr upto +7 oxidation state
Match the statements given in Column I with the oxidation states given in Column II. Column I Column II (i) Oxidation state of Mn in MnO2 is
(a) + 2 (ii) Most stable oxidation state of Mn is
(c) + 4 Mn in oxides is
(b) + 3 (iii) Most stable oxidation state of
(d) + 5 (iv) Characteristic oxidation (e) + 7 state of lanthanoids is
Match the solutions given in Column I and the colours given in Column II. Column I Column II (Aqueous solution of salt) (Colour) (i) FeSO4.7H2O
(a) Green (ii) NiCl2.4H2O
(d) Pale green (v) Cu2Cl2 (e) Pink (f) Colourless
(b) Light pink (iii) MnCl2.4H2O
(c) Blue (iv) CoCl2.6H2O
Match the property given in Column I with the element given in Column II. Column I (Property) Column II (Element) (i) Lanthanoid which shows
(a) Pm +4 oxidation state (ii) Lanthanoid which can show +2
(b) Ce oxidation state (iii) Radioactive lanthanoid
(d) Eu electronic configuration in +3 oxidation state (v) Lanthanoid which has 4f (e) Gd electronic configuration in +3 oxidation state (f) Dy
(c) Lu (iv) Lanthanoid which has 4f
Match the properties given in Column I with the metals given in Column II. Column I (Property) Column II (Metal) (i) Element with highest second
(a) Co ionisation enthalpy (ii) Element with highest third ionisation
(c) Cu (iv) Element with highest heat of atomisation
(b) Cr enthalpy (iii) M in M (CO)6 is
(d) Zn (e) Ni
Assertion (A): Cu2+ iodide is not known. –
Reason (R): Cu2+ oxidises I to iodine.
Assertion (A): Separation of Zr and Hf is difficult.
Reason (R): Because Zr and Hf lie in the same group of the periodic table.
Assertion (A): Actinoids form relatively less stable complexes as compared to lanthanoids.
Reason (R): Actinoids can utilise their 5f orbitals along with 6d orbitals in bonding but lanthanoids do not use their 4f orbital for bonding.
Assertion (A): Cu cannot liberate hydrogen from acids.
Reason (R): Because it has positive electrode potential.
Assertion (A): The highest oxidation state of osmium is +8.
Reason (R): Osmium is a 5d-block element.
Identify A to E and also explain the reactions involved. CuCO3 CuO (D) heat with CuS Ca(OH)2 (A) (E) HNO3(conc.) Milky (B) NH3(aq.) CO2 (C) Ca(HCO3)2 Blue solution Clear solution
A = Cu B = Cu(NO3)2 C = [Cu(NH3)4] D = CO2 E = CaCO3 F = Cu2[Fe(CN)6] G = Ca (HCO3)2 CuCO3 → CuO + CO2 CuO + CuS → Cu + SO2 (A) Cu + 4HNO3 (Conc) → Cu (NO3)2 + 2NO + 2H2O (B) Cu2+ + NH3 → [Cu(NH3)4] (B) (C) Ca(OH)2 + CO2 → CaCO3 + H2O (D) (E) CaCO3 + H2O + CO2 → Ca (HCO3)2
When a chromite ore
(A) is fused with sodium carbonate in free excess of air and the product is dissolved in water, a yellow solution of compound
(B) is obtained. After treatment of this yellow solution with sulphuric acid, compound
(C) can be crystallised from the solution. When compound (C) is treated with KCl, orange crystals of compound
(D) crystallise out. Identify A to D and also explain the reactions.
When an oxide of manganese
(A) is fused with KOH in the presence of an oxidising agent and dissolved in water, it gives a dark green solution of compound
(B) . Compound (B) disproportionates in neutral or acidic solution to give purple compound
(C) . An alkaline solution of compound (C) oxidises potassium iodide solution to a compound
(D) and compound (A) is also formed. Identify compounds A to D and also explain the reactions involved.
On the basis of Lanthanoid contraction, explain the following :
Hint : (i) As the size decreases covalent character increases. Therefore La2O3 is more ionic and Lu2O3 is more covalent. (ii) As the size decreases from La to Lu, stability of oxosalts also decreases. (iii) Stability of complexes increases as the size of lanthanoids decreases. (iv) Radii of 4d and 5d block elements will be almost same. (v) Acidic character of oxides increases from La to Lu.
(a) Answer the following questions :
Mention the type of compounds formed when small atoms like H, C and N get trapped inside the crystal lattice of transition metals. Also give physical and chemical characteristics of these compounds.
Interstitial compounds. Characteristic properties : (i) High melting points, higher than those of pure metals. (ii) Very hard. (iii) Retain metallic conductivity. (iv) Chemically inert.
A violet compound of manganese
(A) decomposes on heating to liberate oxygen and compounds
(B) and
(C) of manganese are formed. Compound (C) reacts with KOH in the presence of potassium nitrate to give compound (B). On heating compound (C) with conc. H2SO4 and NaCl, chlorine gas is liberated and a compound
(D) of manganese along with other products is formed. Identify compounds A to D and also explain the reactions involved. 115 d- and f- Block Elements