How many \(f\) electrons are present in the ground-state electronic configuration of Np (Z = 93)?
1. 18 2. 14
3. 4 4. 21

Subtopic:  f-Block Elements- Properties & Uses |
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Cerium (IV) has a noble gas configuration. Which of the following is correct statement about it ?
1. It will not prefer to undergo redox reactions.
2. It will prefer to gain electron and act as an oxidizing agent
3. It will prefer to give away an electron and behave as reducing agent
4. It acts as both, oxidizing and reducing agent.
Subtopic:  f-Block Elements- Properties & Uses |
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The electronic configuration of Platinum (atomic number: 78) is:

1. \(\mathrm{[Xe] 4f^{14} 5d^9 6s^1}\) 2. \(\mathrm{[Kr] 4f^{14} 5d^{10}}\)
3. \(\mathrm{[Xe] 4f^{14} 5d^{10}}\) 4. \(\mathrm{[Xe] 4f^{14} 5d^8 6s^2}\)
Subtopic:  d-Block Elements- Properties & Uses |
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Given below are two statements.
Statement I: Iron (III) catalyst, acidified \(\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7\) and neutral \(\mathrm{KMnO}_4\) have the ability to oxidise \(​​\mathrm{I}^{-}\)to \(\mathrm{I}_2\) independently.
Statement II: Manganate ion is paramagnetic in nature and involves \(\mathrm{p} \pi-\mathrm{p} \pi\) bonding.
1. Both Statement I and Statement II are true
2. Both Statement I and Statement II are false
3. Statement I is true but Statement II is false
4. Statement I is false but Statement II is true
Subtopic:  Chemistry of Mn and Cr Compounds |
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What is the difference in the spin-only magnetic moment (in B.M.) between manganese (Mn) in KMnO4 and manganese species in the product (X) of the given reaction?
\(\mathrm{KMnO}_4 \xrightarrow{\mathrm{H}^{+}} \mathrm{X} \text { (product having } \mathrm{Mn} \text { ) }\)

1. 2 
2. 4 
3. 0 
4. 6
Subtopic:  d-Block Elements- Properties & Uses |
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The ratio of the number of electrons gained by acidified KMnO4 and acidified K2Cr2O7, respectively, in the given reaction is:
KMnO4 \( \xrightarrow[]{H^{+}}\) Mn2+
K2Cr2O7 \( \xrightarrow[]{H^{+}}\) Cr3+
1. 5:6 2. 6:5
3. 3:5 4. 5:3
Subtopic:  Chemistry of Mn and Cr Compounds |
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Correct statement for the given reaction is: 
\(\mathrm{Cr}_2 \mathrm{O}_7{ }^{2-}(a q)+\mathrm{SO}_3{ }^{2-}(a q) \rightarrow \mathrm{Cr}^{3+}(a q)+\mathrm{SO}_4{ }^{2-}(a q)\) 

1. Generally \(SO_4^{2-}\) is not a reducing agent. 
2. \(Cr^{3+} \) can be further reduced in this reaction. 
3. Oxidation by \(SO_3^{2-} \) gives \(Cr^{3+} \)
4. In \(Cr_2O^{2-}_7 \) oxidation state of Cr is not +6.
Subtopic:  d-Block Elements- Properties & Uses |
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The pair in the actinide series that can have the highest oxidation state is: 

1. Np, Pu 
2. U, Am 
3. U, Cm 
4. Am, Cm 
Subtopic:  d-Block Elements- Properties & Uses |
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When \(V_2O_5\) reacts with base and acid, X  and Y are formed respectively. X and Y are: 
 
1. \(X = VO^{3-}_4 ,~~~Y = VO^{+} _2\)
2. \(X = VO^{3-}_4 ,~~~Y = VO^{+}_4 \)
3. \(X = VO^{+}_4 ,~~~Y = VO^{3-}_4 \)
4. \(X = VO^{2+} ,~~~Y = V_2O_{3} \)
Subtopic:  d-Block Elements- Properties & Uses |
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The following fluoride is not known: 

1. CrF6
2  MnF6
3. MnF4
4. CrF5
Subtopic:  d-Block Elements- Properties & Uses |
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