NEET Test Series from KOTA - 10 Papers In MS WORD
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CHXII08:THE D- & F-BLOCK ELEMENTS
321747
Which of the following oxidation states is common for all lanthanides?
1 \({\rm{ + 2}}\)
2 \({\rm{ + 3}}\)
3 \({\rm{ + 4}}\)
4 \({\rm{ + 5}}\)
Explanation:
Conceptual Questions
KCET - 2018
CHXII08:THE D- & F-BLOCK ELEMENTS
321748
Which of the following acts as a strong reducing agent?(Atomic number: \({\mathrm{\mathrm{Ce}=58, \mathrm{Eu}=63}}\), \({\mathrm{\mathrm{Gd}=64, \mathrm{Lu}=71)}}\)
1 \({\mathrm{\mathrm{Eu}^{2+}}}\)
2 \({\mathrm{\mathrm{Lu}^{3+}}}\)
3 \({\mathrm{\mathrm{Gd}^{3+}}}\)
4 \({\mathrm{\mathrm{Ce}^{4+}}}\)
Explanation:
\({\mathrm{\mathrm{Eu}^{2+}}}\) oxidizes readily to give more stable \({\mathrm{\mathrm{Eu}^{3+}}}\). So, the correct option is (1).
JEE Main - 2024
CHXII08:THE D- & F-BLOCK ELEMENTS
321749
Why Gd has a \(5 \mathrm{~d}^{1}\) arrangement?
1 Reason: Gd shows +2 oxidation state
2 Reason : Poor shielding of 4 f - electrons
3 Reason : High effective nuclear charge
4 Reason : This leaves a half-filled 4 f -level
Explanation:
\(\mathrm{Gd}:[\mathrm{Xe}] 4 \mathrm{f}^{7} 5 \mathrm{~d}^{1} 6 \mathrm{~s}^{2}\). (Due to half-filled f - orbitals).
CHXII08:THE D- & F-BLOCK ELEMENTS
321750
Which of the following statement is correct?
1 \(\mathrm{Ce}^{4+}\) is oxidising agent
2 \(\mathrm{Ce}^{4+}\) is reducing agent
3 \(\mathrm{Ce}^{3+}\) has noble gas configuration
4 Ce has stable configuration
Explanation:
. The \(\mathrm{E}^{\circ}\) value for \(\mathrm{Ce}^{4+} / \mathrm{Ce}^{3+}\) is +1.74 , Therefore, \(\mathrm{Ce}^{4+}\) gains electrons and behaves as an oxidising agent.
321747
Which of the following oxidation states is common for all lanthanides?
1 \({\rm{ + 2}}\)
2 \({\rm{ + 3}}\)
3 \({\rm{ + 4}}\)
4 \({\rm{ + 5}}\)
Explanation:
Conceptual Questions
KCET - 2018
CHXII08:THE D- & F-BLOCK ELEMENTS
321748
Which of the following acts as a strong reducing agent?(Atomic number: \({\mathrm{\mathrm{Ce}=58, \mathrm{Eu}=63}}\), \({\mathrm{\mathrm{Gd}=64, \mathrm{Lu}=71)}}\)
1 \({\mathrm{\mathrm{Eu}^{2+}}}\)
2 \({\mathrm{\mathrm{Lu}^{3+}}}\)
3 \({\mathrm{\mathrm{Gd}^{3+}}}\)
4 \({\mathrm{\mathrm{Ce}^{4+}}}\)
Explanation:
\({\mathrm{\mathrm{Eu}^{2+}}}\) oxidizes readily to give more stable \({\mathrm{\mathrm{Eu}^{3+}}}\). So, the correct option is (1).
JEE Main - 2024
CHXII08:THE D- & F-BLOCK ELEMENTS
321749
Why Gd has a \(5 \mathrm{~d}^{1}\) arrangement?
1 Reason: Gd shows +2 oxidation state
2 Reason : Poor shielding of 4 f - electrons
3 Reason : High effective nuclear charge
4 Reason : This leaves a half-filled 4 f -level
Explanation:
\(\mathrm{Gd}:[\mathrm{Xe}] 4 \mathrm{f}^{7} 5 \mathrm{~d}^{1} 6 \mathrm{~s}^{2}\). (Due to half-filled f - orbitals).
CHXII08:THE D- & F-BLOCK ELEMENTS
321750
Which of the following statement is correct?
1 \(\mathrm{Ce}^{4+}\) is oxidising agent
2 \(\mathrm{Ce}^{4+}\) is reducing agent
3 \(\mathrm{Ce}^{3+}\) has noble gas configuration
4 Ce has stable configuration
Explanation:
. The \(\mathrm{E}^{\circ}\) value for \(\mathrm{Ce}^{4+} / \mathrm{Ce}^{3+}\) is +1.74 , Therefore, \(\mathrm{Ce}^{4+}\) gains electrons and behaves as an oxidising agent.
321747
Which of the following oxidation states is common for all lanthanides?
1 \({\rm{ + 2}}\)
2 \({\rm{ + 3}}\)
3 \({\rm{ + 4}}\)
4 \({\rm{ + 5}}\)
Explanation:
Conceptual Questions
KCET - 2018
CHXII08:THE D- & F-BLOCK ELEMENTS
321748
Which of the following acts as a strong reducing agent?(Atomic number: \({\mathrm{\mathrm{Ce}=58, \mathrm{Eu}=63}}\), \({\mathrm{\mathrm{Gd}=64, \mathrm{Lu}=71)}}\)
1 \({\mathrm{\mathrm{Eu}^{2+}}}\)
2 \({\mathrm{\mathrm{Lu}^{3+}}}\)
3 \({\mathrm{\mathrm{Gd}^{3+}}}\)
4 \({\mathrm{\mathrm{Ce}^{4+}}}\)
Explanation:
\({\mathrm{\mathrm{Eu}^{2+}}}\) oxidizes readily to give more stable \({\mathrm{\mathrm{Eu}^{3+}}}\). So, the correct option is (1).
JEE Main - 2024
CHXII08:THE D- & F-BLOCK ELEMENTS
321749
Why Gd has a \(5 \mathrm{~d}^{1}\) arrangement?
1 Reason: Gd shows +2 oxidation state
2 Reason : Poor shielding of 4 f - electrons
3 Reason : High effective nuclear charge
4 Reason : This leaves a half-filled 4 f -level
Explanation:
\(\mathrm{Gd}:[\mathrm{Xe}] 4 \mathrm{f}^{7} 5 \mathrm{~d}^{1} 6 \mathrm{~s}^{2}\). (Due to half-filled f - orbitals).
CHXII08:THE D- & F-BLOCK ELEMENTS
321750
Which of the following statement is correct?
1 \(\mathrm{Ce}^{4+}\) is oxidising agent
2 \(\mathrm{Ce}^{4+}\) is reducing agent
3 \(\mathrm{Ce}^{3+}\) has noble gas configuration
4 Ce has stable configuration
Explanation:
. The \(\mathrm{E}^{\circ}\) value for \(\mathrm{Ce}^{4+} / \mathrm{Ce}^{3+}\) is +1.74 , Therefore, \(\mathrm{Ce}^{4+}\) gains electrons and behaves as an oxidising agent.
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
CHXII08:THE D- & F-BLOCK ELEMENTS
321747
Which of the following oxidation states is common for all lanthanides?
1 \({\rm{ + 2}}\)
2 \({\rm{ + 3}}\)
3 \({\rm{ + 4}}\)
4 \({\rm{ + 5}}\)
Explanation:
Conceptual Questions
KCET - 2018
CHXII08:THE D- & F-BLOCK ELEMENTS
321748
Which of the following acts as a strong reducing agent?(Atomic number: \({\mathrm{\mathrm{Ce}=58, \mathrm{Eu}=63}}\), \({\mathrm{\mathrm{Gd}=64, \mathrm{Lu}=71)}}\)
1 \({\mathrm{\mathrm{Eu}^{2+}}}\)
2 \({\mathrm{\mathrm{Lu}^{3+}}}\)
3 \({\mathrm{\mathrm{Gd}^{3+}}}\)
4 \({\mathrm{\mathrm{Ce}^{4+}}}\)
Explanation:
\({\mathrm{\mathrm{Eu}^{2+}}}\) oxidizes readily to give more stable \({\mathrm{\mathrm{Eu}^{3+}}}\). So, the correct option is (1).
JEE Main - 2024
CHXII08:THE D- & F-BLOCK ELEMENTS
321749
Why Gd has a \(5 \mathrm{~d}^{1}\) arrangement?
1 Reason: Gd shows +2 oxidation state
2 Reason : Poor shielding of 4 f - electrons
3 Reason : High effective nuclear charge
4 Reason : This leaves a half-filled 4 f -level
Explanation:
\(\mathrm{Gd}:[\mathrm{Xe}] 4 \mathrm{f}^{7} 5 \mathrm{~d}^{1} 6 \mathrm{~s}^{2}\). (Due to half-filled f - orbitals).
CHXII08:THE D- & F-BLOCK ELEMENTS
321750
Which of the following statement is correct?
1 \(\mathrm{Ce}^{4+}\) is oxidising agent
2 \(\mathrm{Ce}^{4+}\) is reducing agent
3 \(\mathrm{Ce}^{3+}\) has noble gas configuration
4 Ce has stable configuration
Explanation:
. The \(\mathrm{E}^{\circ}\) value for \(\mathrm{Ce}^{4+} / \mathrm{Ce}^{3+}\) is +1.74 , Therefore, \(\mathrm{Ce}^{4+}\) gains electrons and behaves as an oxidising agent.