162432
Among the following transition elements, pick out the element/elements with highest second ionisation energy:
1 \(\mathrm{V}(Z=23)\)
2 \(\operatorname{Cr}(Z=24)\)
3 \(\mathrm{Mn}(Z=25)\)
4 \(\mathrm{Cu}(Z=29)\)
Explanation:
Correct order of second ionisation energy \(\mathrm{V}<\mathrm{Mn}<\mathrm{Cr}<\mathrm{Cu}\)
NCERT-XII-95
9 RBTS PAPER
162433
\(\mathrm{KMnO}_4\) is a strong oxidising agent in acid medium. To provide an acid medium \(\mathrm{H}_2 \mathrm{SO}_4\) is used instead of \(\mathrm{HCl}\). This is because:
1 \(\mathrm{H}_2 \mathrm{SO}_4\) is stronger acid than \(\mathrm{HCl}\)
2 \(\mathrm{HCl}\) is oxidised by \(\mathrm{KMnO}_4\) to \(\mathrm{Cl}_2\)
3 \(\mathrm{H}_2 \mathrm{SO}_4\) is dibasic acid
4 rate is faster in presence of \(\mathrm{H}_2 \mathrm{SO}_4\)
Explanation:
because \(\mathrm{HCl}\) is oxidised by \(\mathrm{KMnO}_4\) to \(\mathrm{Cl}_2\)
NCERT-XII-108
9 RBTS PAPER
162434
Acidified \(\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7\) turns green when \(\mathrm{Na}_2 \mathrm{SO}_3\) is added to it. This is due to the formation of:
1 \(\mathrm{Cr}_2\left(\mathrm{SO}_4\right)_3\)
2 \(\mathrm{CrO}_4^{2-}\)
3 \(\mathrm{CrSO}_4\)
4 None of these
Explanation:
NCERT-XII-106
9 RBTS PAPER
162435
Oxidation number and EAN of \(\mathrm{Fe}\) in \(\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}\) are respectively :
162432
Among the following transition elements, pick out the element/elements with highest second ionisation energy:
1 \(\mathrm{V}(Z=23)\)
2 \(\operatorname{Cr}(Z=24)\)
3 \(\mathrm{Mn}(Z=25)\)
4 \(\mathrm{Cu}(Z=29)\)
Explanation:
Correct order of second ionisation energy \(\mathrm{V}<\mathrm{Mn}<\mathrm{Cr}<\mathrm{Cu}\)
NCERT-XII-95
9 RBTS PAPER
162433
\(\mathrm{KMnO}_4\) is a strong oxidising agent in acid medium. To provide an acid medium \(\mathrm{H}_2 \mathrm{SO}_4\) is used instead of \(\mathrm{HCl}\). This is because:
1 \(\mathrm{H}_2 \mathrm{SO}_4\) is stronger acid than \(\mathrm{HCl}\)
2 \(\mathrm{HCl}\) is oxidised by \(\mathrm{KMnO}_4\) to \(\mathrm{Cl}_2\)
3 \(\mathrm{H}_2 \mathrm{SO}_4\) is dibasic acid
4 rate is faster in presence of \(\mathrm{H}_2 \mathrm{SO}_4\)
Explanation:
because \(\mathrm{HCl}\) is oxidised by \(\mathrm{KMnO}_4\) to \(\mathrm{Cl}_2\)
NCERT-XII-108
9 RBTS PAPER
162434
Acidified \(\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7\) turns green when \(\mathrm{Na}_2 \mathrm{SO}_3\) is added to it. This is due to the formation of:
1 \(\mathrm{Cr}_2\left(\mathrm{SO}_4\right)_3\)
2 \(\mathrm{CrO}_4^{2-}\)
3 \(\mathrm{CrSO}_4\)
4 None of these
Explanation:
NCERT-XII-106
9 RBTS PAPER
162435
Oxidation number and EAN of \(\mathrm{Fe}\) in \(\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}\) are respectively :
162432
Among the following transition elements, pick out the element/elements with highest second ionisation energy:
1 \(\mathrm{V}(Z=23)\)
2 \(\operatorname{Cr}(Z=24)\)
3 \(\mathrm{Mn}(Z=25)\)
4 \(\mathrm{Cu}(Z=29)\)
Explanation:
Correct order of second ionisation energy \(\mathrm{V}<\mathrm{Mn}<\mathrm{Cr}<\mathrm{Cu}\)
NCERT-XII-95
9 RBTS PAPER
162433
\(\mathrm{KMnO}_4\) is a strong oxidising agent in acid medium. To provide an acid medium \(\mathrm{H}_2 \mathrm{SO}_4\) is used instead of \(\mathrm{HCl}\). This is because:
1 \(\mathrm{H}_2 \mathrm{SO}_4\) is stronger acid than \(\mathrm{HCl}\)
2 \(\mathrm{HCl}\) is oxidised by \(\mathrm{KMnO}_4\) to \(\mathrm{Cl}_2\)
3 \(\mathrm{H}_2 \mathrm{SO}_4\) is dibasic acid
4 rate is faster in presence of \(\mathrm{H}_2 \mathrm{SO}_4\)
Explanation:
because \(\mathrm{HCl}\) is oxidised by \(\mathrm{KMnO}_4\) to \(\mathrm{Cl}_2\)
NCERT-XII-108
9 RBTS PAPER
162434
Acidified \(\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7\) turns green when \(\mathrm{Na}_2 \mathrm{SO}_3\) is added to it. This is due to the formation of:
1 \(\mathrm{Cr}_2\left(\mathrm{SO}_4\right)_3\)
2 \(\mathrm{CrO}_4^{2-}\)
3 \(\mathrm{CrSO}_4\)
4 None of these
Explanation:
NCERT-XII-106
9 RBTS PAPER
162435
Oxidation number and EAN of \(\mathrm{Fe}\) in \(\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}\) are respectively :
162432
Among the following transition elements, pick out the element/elements with highest second ionisation energy:
1 \(\mathrm{V}(Z=23)\)
2 \(\operatorname{Cr}(Z=24)\)
3 \(\mathrm{Mn}(Z=25)\)
4 \(\mathrm{Cu}(Z=29)\)
Explanation:
Correct order of second ionisation energy \(\mathrm{V}<\mathrm{Mn}<\mathrm{Cr}<\mathrm{Cu}\)
NCERT-XII-95
9 RBTS PAPER
162433
\(\mathrm{KMnO}_4\) is a strong oxidising agent in acid medium. To provide an acid medium \(\mathrm{H}_2 \mathrm{SO}_4\) is used instead of \(\mathrm{HCl}\). This is because:
1 \(\mathrm{H}_2 \mathrm{SO}_4\) is stronger acid than \(\mathrm{HCl}\)
2 \(\mathrm{HCl}\) is oxidised by \(\mathrm{KMnO}_4\) to \(\mathrm{Cl}_2\)
3 \(\mathrm{H}_2 \mathrm{SO}_4\) is dibasic acid
4 rate is faster in presence of \(\mathrm{H}_2 \mathrm{SO}_4\)
Explanation:
because \(\mathrm{HCl}\) is oxidised by \(\mathrm{KMnO}_4\) to \(\mathrm{Cl}_2\)
NCERT-XII-108
9 RBTS PAPER
162434
Acidified \(\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7\) turns green when \(\mathrm{Na}_2 \mathrm{SO}_3\) is added to it. This is due to the formation of:
1 \(\mathrm{Cr}_2\left(\mathrm{SO}_4\right)_3\)
2 \(\mathrm{CrO}_4^{2-}\)
3 \(\mathrm{CrSO}_4\)
4 None of these
Explanation:
NCERT-XII-106
9 RBTS PAPER
162435
Oxidation number and EAN of \(\mathrm{Fe}\) in \(\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}\) are respectively :
