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Question 31 Mark
Write down the electronic configuration of:

Lu2+

Answer
Lu2+: 1s2 2s2 2p3s3p3d10 4s2 4p6 4d10 5s5p6 4f 14 5d1
Or, [Xe]54 2f14 3d3
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Question 41 Mark
Describe the oxidising action of potassium dichromate and write the ionic equations for its reaction with:
Iodide
Answer
K2Cr2Ois a powerful oxidising agent. In dilute sulphuric acid medium the oxidation state of Cr changes from +6 to +3. The oxidising action can be represented as follows:
$\text{Cr}_2\text{O}^{2-}_7+14\text{H}^++6\text{e}^-\rightarrow2\text{Cr}^{3+}+7\text{H}_2\text{O}$
Iodide: Iodide ion (J-) is oxidised to I by the acidfied solution of K2Cr2O7.
Reaction:
$\text{Cr}_2\text{O}_7^{2-}+14\text{H}^++6\text{e}^-\rightarrow2\text{Cr}^{3+}+7\text{H}_2\text{O}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 6\text{I}^-\rightarrow3\text{I}_2+6\text{e}^-\\ \overline{\text{Cr}_2\text{O}^{2-}_7+14\text{H}^+6\text{I}^-\rightarrow3\text{I}_2+2\text{Cr}^{3+}+7\text{H}_2\text{O}}$
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Question 51 Mark
Calculate the ‘spin only’ magnetic moment of M2+(aq) ion (Z = 27).
Answer
Atomic number (27) = [Ar] 3d74s2
M2+ = [Ar] 3d7
Thus it has three unpaired electrons
therefore magnetic moment is
$\mu=\sqrt{\text{n}(\text{n}+2)}$
Where n = total number of unpaired electron
$\mu=\sqrt{3(3+2)}$
$\mu=\sqrt{15}$
$\mu=3.8\text{M}$
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Question 61 Mark
In the series Sc (Z = 21) to Zn (Z = 30), the enthalpy of atomisation of zinc is the lowest, i.e., 126 kJ mol–1. Why?
Answer
The extent of metallic bonding an element undergoes deideds the enthalpy of atomization the more extensive the metallic bonding of an element the more will be its enthalpy of atomization.
Sc & Zn belongs to 3rd group of periodic table. In all transition metals (except Zn, electronic configuration: 3d10 4s2), there are some unpaired electrons that account for their stronger metallic bonding. Due to the absence of these unpaired electrons, the inter-atomic electronic bonding is the weakest in Zn and as a result, it has the least enthalpy of atomization.
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Question 71 Mark
Describe the oxidising action of potassium dichromate and write the ionic equations for its reaction with:
iron(II) solution
Answer
Iron (II) solution: Ferrous salts (Fe2+) are oxidised ferric (Fe3+) salts when they are treated with acidified K2Cr2O7.
Reaction:
$\text{Cr}_2\text{O}_7^{2-}+14\text{H}^{+}+6\text{e}^-\rightarrow2\text{Cr}^{3+}+7\text{H}_2\text{O}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 6\text{Fe}^{2+}\rightarrow6\text{Fe}^{3+}+6\text{e}^-\\\overline{\text{Cr}_2\text{O}^{2-}_7+6\text{Fe}^2+14\text{H}^+\rightarrow2\text{Cr}^{3+}+6\text{Fe}^{3+}+7\text{H}_2\text{O}}$
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Question 81 Mark
Explain why Cu+ ion is not stable in aqueous solutions?
Answer
In an aqueous medium, Cu2+ is more stable than Cu+. This is because although energy is required to remove one electron from Cu+ to Cu2+, high hydration energy of Cu2+ compensates for it. Therefore, Cu+ ion in an aqueous solution is unstable. It disproportionate to give Cu2+ and Cu.
$2\text{Cu}^+_{(\text{aq})}\rightarrow\text{Cu}^+_{(\text{aq})}+\text{Cu}_{(\text{s})}$
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Question 91 Mark
Describe the oxidising action of potassium dichromate and write the ionic equations for its reaction with:
H2S
Answer
H2S: H2S is oxidised to sulphur.
$\text{Cr}_2\text{O}_7^{2-}+3\text{H}_2\text{S}+8\text{H}^+\rightarrow2\text{Cr}^{3+}+7\text{H}_2\text{O}+3\text{S}$
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Question 101 Mark
Which of the 3d series of the transition metals exhibits the largest number of oxidation states and why?
Answer
Mn (Z = 25) = 3d54s2
Mn has the maximum number of unpaired electrons present in the d-subshell (5 electrons). Hence, Mn exhibits the largest number of oxidation states, ranging from +2 to +7.
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Question 111 Mark
Silver atom has completely filled d orbitals (4d10) in its ground state. How can you say that it is a transition element?
Answer
The outer electronic configuration of Ag (Z = 47) is 4d105s1. It shows +1 and +2O.S. (in AgO and AgF2). And in +2O.S. , the electronic configuration is d9 i.e., d - subshell is incompletely filled. Hence, it is a transition element.
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Question 121 Mark
Compare the chemistry of actinoids with that of the lanthanoids with special reference to:
Electronic configuration
Answer
Electronic configuration: The general electronic configuration of lanthanoids is [Xe]54 4f1-14 5d0·1 6s2 and that of actinoids is [Rn]86 5f0-14 6d0-1 7s2, lanthanoids belong to 4f series whereas actinoids belong to 5f-series.
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Question 131 Mark
Which is the last element in the series of the actinoids? Write the electronic configuration of this element. Comment on the possible oxidation state of this element.
Answer
Lawrencium (Lr) is the last element in the series of antinoids.
Its electronic configuration is [Rn] 5f14 6d1 7s2. The possible oxidation state of Lawrencium is +3.
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Question 151 Mark
Use Hund’s rule to derive the electronic configuration of Ce3+ ion, and calculate its magnetic moment on the basis of ‘spin-only’ formula.
Answer
Ce: 1s2 2s3p3s3p3d10 4s2 4p6 4d10 5s5p6 4f5d6s2
Magnetic moment can be calculated as:
$\mu=\sqrt{\text{n}(\text{n}+2)}$
Where,
n = number of unpaired electrons
The electronic configuration of Ce3+: 1s2 2s2 2p3s3p3d10 4s2 4p6 4d10 5s5p6 4f1
In Ce3+, n = 1
Therefore, $\mu=\sqrt{2(2+2)}$
$=\mu=\sqrt{2\times4}$
$=\mu=\sqrt{8}$
$=\mu=2\sqrt{2}$
= 2.828 BM
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Question 161 Mark
The $\text{E}^\ominus(\text{M}^{2+}/\text{M})$ value for copper is positive (+0.34V). What is possibly the reason for this? (Hint: consider its high $\Delta_\text{a}\text{H}^\ominus$ and low $\Delta_\text{hyd}\text{H}^\ominus$).
Answer
$\text{E}^\ominus(\text{M}^{2+}/\text{M})$ for any metal is related to the sum of enthelpy changes taking place in following steps:
$\text{M(s)}+\Delta_\text{a}\text{H}\rightarrow\text{M(g)}$
$\text{M(g)}+\Delta_\text{i}\text{H}\rightarrow\text{M}^{2+}{\text{(g)}}$
$\text{M(g)}+\Delta_\text{i}\text{H}\rightarrow\text{M}^{2+}{\text{(g)}}$
Cu has a high enthelpy of atomisation $(\Delta_\text{a}\text{H})$ and a low enthelpy of hydration $(\Delta_\text{hyd}\text{H})$. The high energy required to transform Cu(s) to Cu2+(aq)  is not balenced by its hydration enthelpy. Hence $\text{E}^\ominus(\text{Cu}^{2+}/\text{Cu})$ is positive.
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Question 171 Mark
Actinoid contraction is greater from element to element than lanthanoid contraction. Why?
Answer
This is due to poor shielding by 5f-electrons in the actinoids than that by 4f e-1s in lanthanoids.
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Question 181 Mark
Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only?
Answer
Both oxide and fluoride ions are highly electronegative and have a very small size. Due to these properties, they are able to oxidize the metal to its highest oxidation state.
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Question 191 Mark
Write down the electronic configuration of:

Pm3+

Answer
Pm3+: 1s2 2s2p3s3p3d10 4s2 4p6 4d10 5s5p6 4f4
Or, [Xe]54 3d3
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Question 201 Mark
Write down the electronic configuration of:

Ce4+

Answer
Ce4+: 1s2 2s2 2p3s3p3d10 4s2 4p6 4d10 5s5p6
Or, [Xe]54
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Question 211 Mark
Which is a stronger reducing agent Cr2+ or Fe2+ and why?
Answer
Cr2+ is a stronger reducing agent than Fe2+. This is because E°(Cr3+/Cr2+) is negative (-0.41 V) whereas E°(Fe3+/Fe2+) is positive (+ 0.77 V). Thus, Cr2+ is easily oxidised to Fe3+ but Fe2+ cannot be easily oxidised to Fe3+.
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Question 221 Mark
Write down the electronic configuration of:
Th4+
Answer
Th4+ = 1s2, 2s2, 2p6, 3s2, 3p6, 3d10, 4s2, 4p6, 4d10, 4f14, 5s2, 5p6, 5d10, 6s2, 6p6.
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Question 231 Mark
What are the different oxidation states exhibited by the lanthanoids?
Answer
In the lanthanide series, +3 oxidation state is most common i.e., Ln(III) compounds are predominant. However, +2 and +4 oxidation states can also be found in the solution or in solid compounds.
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Question 251 Mark
Why do transition elements show similarities along the horizontal period?
Answer
All transition elements contain incompletely filled d-subshell whereas the outer shell electronic configuration remains the same.
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Question 281 Mark
Electronic configuration of a transition element X in +3 oxidation state is [Ar]3d5. What is its atomic number?
  1. 25
  2. 26
  3. 27
  4. 24
Answer
  1. 26

Explanation:

Electronic configuration of element X = [Ar] 3d5+ oxidation state = 18 + 5 + 3 = 26.

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Question 291 Mark
Arrange the following in increasing order of acidic character: CrO3, CrO, Cr2O3.
Answer
CrO < Cr2O3 < CrO3. Higher the oxidation state, more will be acidic character.
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Question 301 Mark
A transition metal which does not exhibit variation in oxidation state in its compounds.
Answer
Scandium (Sc).
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Question 311 Mark
Why does a transition series contain 10 elements?
Answer
There are five d-orbitals in an energy level and each orbital can contain two electrons. As we move from one element to the next, an electron is added and for complete filling of the five d-orbitals, 10 electrons are required.
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Question 331 Mark
Name a transition element which does not exhibit variable oxidation state
Answer
Scandium (Z = 21) does not exhibit variable oxidation states.
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Question 341 Mark
On the basis of lanthanoid contraction, explain the following:
Trends in the stability of oxo salts of lanthanoids from La to Lu.
Answer
As the size decreases from La to Lu, stability of oxosalts also decreases.
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Question 351 Mark
Fe has higher melting point than Cu. Why?
Answer
This is because Fe (3d6 4s1) has four unpaired electrons in 3d-subshell. While Cu (3d10 4s1) has only one electron in the 4s-subshell. Hence, metallic bonds in Fe are much stronger than those in Cu.
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Question 361 Mark
Write any one use of pyrophoric alloys.
Answer
Pyrophoric alloys emit sparks when struck. Hence, they are used in making flints for lighters.
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Question 371 Mark
On the basis of lanthanoid contraction, explain the following:
Radii of 4d and 5d block elements.
Answer
Radii of 4d and 5d block elements will be almost same.
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Question 381 Mark
Why are transition elements so named?
Answer
Transition elements are so named because their properties are in between those of s-block and p-block.
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Question 391 Mark
What is meant by 'lanthanoid contraction'?
Answer
The regular decrease in the atomic and ionic radii/(having the same charge) of Lanthanoids with increasing atomic number is known as Lanthanoid contraction.
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Question 401 Mark
Answer the following questions:
Which element of the first transition series has lowest enthalpy ofatomisation?
Answer
because of the completely filled 3d sub shell no unpaired electron is left for metallic bonding.
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Question 411 Mark
On the basis of Lanthanoid contraction, explain the following:
Stability of the complexes of lanthanoids.
Answer
Stability of complexes from La to Lu, increases as the size of the central atom decreases.
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Question 421 Mark
What is the common oxidation state of Cu, Ag and Au?
Answer
The common oxidation state of Cu, Ag and Au is +1.
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Question 431 Mark
Why are transition elements known as d-block elements?
Answer
The last electron enters (n - 1) d-orbital, i.e., d-orbital of the penultimate shell. Hence, these are known as d-block elements.
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Question 451 Mark
On the basis of Lanthanoid contraction, explain the following:
Trends in acidic character of lanthanoid oxides.
Answer
The acidic nature of lanthanoid oxides increases from La to Lu.
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Question 471 Mark
Which divalent metal ion has maximum paramagnetic character among the first transition metals? Why?
Answer
Mn2+ has the maximum paramagnetic character because of the maximum number of unpaired electrons, viz., 5.
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Question 481 Mark
Write the formula of an oxo-anion of Chromium (Cr) in which it shows the oxidation state equal to its group number.
Answer
Formula of oxo-anion of chromium (Cr) in which it shows the oxidation state equal to its group number (6) is $\text{Cr}_2\text{O}^2_{−7}.$
2Cr + (-2 ×7) = -2
2Cr - 14 = -2
2Cr = 12
Cr = +6
Oxidation of Cr in $\text{Cr}_2\text{O}^2_{−7}$ is +6 which is equal to its group number 6.
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Question 491 Mark
Discuss the relative stability in aqueous solutions of +2 oxidation state among the elements: Cr, Mn, Fe and Co. How would you justify this situation?
(At. Nos. Cr = 24, Mn = 25, Fe = 26, Co = 27)
Answer
On the basis of electrochemical series, the standard electrode potential shows the following order:
E°Mn2+/Mn < E°Cr2+/Cr < E°Fe2+/Fe < E°CO2+/CO
Therefore, Co2+ gets easily reduced to metallic cobalt while it is difficult to reduce Mn2+. Hence, Mn2+ will be the most stable and the increasing stability order will be
Co2+ < Fe2+ < Cr2+ < Mn2+
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Question 501 Mark
Write the formula of an oxo-anion of Manganese (Mn) in which it shows the oxidation state equal to its group number.
Answer
MnO4/ KMnO4
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1 Marks Question - Chemistry STD 12 Science Questions - Vidyadip