238924
According to Bohr's theory, the angular momentum for an electron in $5^{\text {th }}$ orbit is
1 $2.5 \mathrm{~h} / \pi$
2 $5 \mathrm{~h} / \pi$
3 $25 \mathrm{~h} / \pi$
4 $5 \pi / 2 \mathrm{~h}$
Explanation:
The angular momentum $(\ell)$ of an electron in a Bohr's orbit is given as $\mathrm{L}=\frac{\mathrm{nh}}{2 \pi}$ It is an integral multiple of $\frac{\mathrm{h}}{2 \pi}$ In the fifth Bohr orbit, the angular momentum of electron is $(\mathrm{L})=\frac{5 \mathrm{~h}}{2 \pi}$ $\begin{aligned} & \mathrm{L}=\frac{2.5 \mathrm{~h}}{\pi} \\ & =2.5 \mathrm{~h} \backslash \pi \end{aligned}$
UPTU/UPSEE-2010
Structure of Atom
238911
Assertion: $d^5$ configuration is more stable than $\mathrm{d}^4$ Reason: $d^5$ has more exchange energy as compared to $\mathrm{d}^4$ because $10 \& 6$ exchanges are possible in $\mathrm{d}^5 \& \mathrm{~d}^4$ respectively.
1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.
3 If Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
: $d^5$ configuration has exactly haif-filled configuration so has more exchange energy. While $\mathrm{d}^4$ has only four such (unpaired) electron these has less exchange energy. Maximum exchange energy leads to the stabilization of the atoms. Therefore, $d^5$ configuration is more stable them that of $\mathrm{d}^4$ configuration.
AIIMS 26 May 2019 (Morning)
Structure of Atom
238918
If an element with atomic number $Z=115$ has been discovered today in which of the following family would it had been placed and identify its electronic configuration. #[Qdiff: Very Easy, QCat: Theory Based, examname: Shift-I , AP- EAPCET-07-09-2021]#
Atomic number $\mathrm{Z}=115$ is Moscovium (Mc) is a synthetic chemical element and an extremely radioactive element. It belongs to nitrogen family and electronic configuration is $115^{\mathrm{Mc}}=[\mathrm{Rn}] 5 \mathrm{f}^{14} 6 \mathrm{~d}^{10} 75^2 7 \mathrm{p}^3$
Structure of Atom
238926
The electrons indentified by quantum numbers n and $l$ (i) $\mathrm{n}=4, l=1 ;$ (ii) $\mathrm{n}=4, l=0$; (iii) $\mathrm{n}=3$, $l=2 ;$ (iv) $\mathrm{n}=3, l=1$ can be placed in order of increasing energy from the lowest to highest as
1 (iv) $<$ (ii) $<$ (iii) $<$ (i)
2 (ii) $<$ (iv) $<$ (i) $<$ (iii)
3 (i) $<$ (iii) $<$ (ii) $<$ (iv)
4 (iii) $<$ (i) $<$ (iv) $<$ (ii)
Explanation:
The greater is the value of $(\mathrm{n}+1)$, the greater si the energy of orbitals. (i) $\mathrm{n}=4, \mathrm{l}=1=4 \mathrm{p}$ orbital (ii) $\mathrm{n}=4, \mathrm{l}=0=4$ s orbital (iii) $\mathrm{n}=3, \mathrm{l}=2=3 \mathrm{~d}$ orbital (iv) $\mathrm{n}=3, \mathrm{l}=1=3 \mathrm{p}$ orbital According to the Aufbau principle, energies of abovementioned orbitals are in the order of - The increasing order of energy (iv) $3 \mathrm{p}<$ (ii) $4 \mathrm{~s}<$ (iii) $3 \mathrm{~s}<$ (i) $4 \mathrm{p}$
238924
According to Bohr's theory, the angular momentum for an electron in $5^{\text {th }}$ orbit is
1 $2.5 \mathrm{~h} / \pi$
2 $5 \mathrm{~h} / \pi$
3 $25 \mathrm{~h} / \pi$
4 $5 \pi / 2 \mathrm{~h}$
Explanation:
The angular momentum $(\ell)$ of an electron in a Bohr's orbit is given as $\mathrm{L}=\frac{\mathrm{nh}}{2 \pi}$ It is an integral multiple of $\frac{\mathrm{h}}{2 \pi}$ In the fifth Bohr orbit, the angular momentum of electron is $(\mathrm{L})=\frac{5 \mathrm{~h}}{2 \pi}$ $\begin{aligned} & \mathrm{L}=\frac{2.5 \mathrm{~h}}{\pi} \\ & =2.5 \mathrm{~h} \backslash \pi \end{aligned}$
UPTU/UPSEE-2010
Structure of Atom
238911
Assertion: $d^5$ configuration is more stable than $\mathrm{d}^4$ Reason: $d^5$ has more exchange energy as compared to $\mathrm{d}^4$ because $10 \& 6$ exchanges are possible in $\mathrm{d}^5 \& \mathrm{~d}^4$ respectively.
1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.
3 If Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
: $d^5$ configuration has exactly haif-filled configuration so has more exchange energy. While $\mathrm{d}^4$ has only four such (unpaired) electron these has less exchange energy. Maximum exchange energy leads to the stabilization of the atoms. Therefore, $d^5$ configuration is more stable them that of $\mathrm{d}^4$ configuration.
AIIMS 26 May 2019 (Morning)
Structure of Atom
238918
If an element with atomic number $Z=115$ has been discovered today in which of the following family would it had been placed and identify its electronic configuration. #[Qdiff: Very Easy, QCat: Theory Based, examname: Shift-I , AP- EAPCET-07-09-2021]#
Atomic number $\mathrm{Z}=115$ is Moscovium (Mc) is a synthetic chemical element and an extremely radioactive element. It belongs to nitrogen family and electronic configuration is $115^{\mathrm{Mc}}=[\mathrm{Rn}] 5 \mathrm{f}^{14} 6 \mathrm{~d}^{10} 75^2 7 \mathrm{p}^3$
Structure of Atom
238926
The electrons indentified by quantum numbers n and $l$ (i) $\mathrm{n}=4, l=1 ;$ (ii) $\mathrm{n}=4, l=0$; (iii) $\mathrm{n}=3$, $l=2 ;$ (iv) $\mathrm{n}=3, l=1$ can be placed in order of increasing energy from the lowest to highest as
1 (iv) $<$ (ii) $<$ (iii) $<$ (i)
2 (ii) $<$ (iv) $<$ (i) $<$ (iii)
3 (i) $<$ (iii) $<$ (ii) $<$ (iv)
4 (iii) $<$ (i) $<$ (iv) $<$ (ii)
Explanation:
The greater is the value of $(\mathrm{n}+1)$, the greater si the energy of orbitals. (i) $\mathrm{n}=4, \mathrm{l}=1=4 \mathrm{p}$ orbital (ii) $\mathrm{n}=4, \mathrm{l}=0=4$ s orbital (iii) $\mathrm{n}=3, \mathrm{l}=2=3 \mathrm{~d}$ orbital (iv) $\mathrm{n}=3, \mathrm{l}=1=3 \mathrm{p}$ orbital According to the Aufbau principle, energies of abovementioned orbitals are in the order of - The increasing order of energy (iv) $3 \mathrm{p}<$ (ii) $4 \mathrm{~s}<$ (iii) $3 \mathrm{~s}<$ (i) $4 \mathrm{p}$
238924
According to Bohr's theory, the angular momentum for an electron in $5^{\text {th }}$ orbit is
1 $2.5 \mathrm{~h} / \pi$
2 $5 \mathrm{~h} / \pi$
3 $25 \mathrm{~h} / \pi$
4 $5 \pi / 2 \mathrm{~h}$
Explanation:
The angular momentum $(\ell)$ of an electron in a Bohr's orbit is given as $\mathrm{L}=\frac{\mathrm{nh}}{2 \pi}$ It is an integral multiple of $\frac{\mathrm{h}}{2 \pi}$ In the fifth Bohr orbit, the angular momentum of electron is $(\mathrm{L})=\frac{5 \mathrm{~h}}{2 \pi}$ $\begin{aligned} & \mathrm{L}=\frac{2.5 \mathrm{~h}}{\pi} \\ & =2.5 \mathrm{~h} \backslash \pi \end{aligned}$
UPTU/UPSEE-2010
Structure of Atom
238911
Assertion: $d^5$ configuration is more stable than $\mathrm{d}^4$ Reason: $d^5$ has more exchange energy as compared to $\mathrm{d}^4$ because $10 \& 6$ exchanges are possible in $\mathrm{d}^5 \& \mathrm{~d}^4$ respectively.
1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.
3 If Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
: $d^5$ configuration has exactly haif-filled configuration so has more exchange energy. While $\mathrm{d}^4$ has only four such (unpaired) electron these has less exchange energy. Maximum exchange energy leads to the stabilization of the atoms. Therefore, $d^5$ configuration is more stable them that of $\mathrm{d}^4$ configuration.
AIIMS 26 May 2019 (Morning)
Structure of Atom
238918
If an element with atomic number $Z=115$ has been discovered today in which of the following family would it had been placed and identify its electronic configuration. #[Qdiff: Very Easy, QCat: Theory Based, examname: Shift-I , AP- EAPCET-07-09-2021]#
Atomic number $\mathrm{Z}=115$ is Moscovium (Mc) is a synthetic chemical element and an extremely radioactive element. It belongs to nitrogen family and electronic configuration is $115^{\mathrm{Mc}}=[\mathrm{Rn}] 5 \mathrm{f}^{14} 6 \mathrm{~d}^{10} 75^2 7 \mathrm{p}^3$
Structure of Atom
238926
The electrons indentified by quantum numbers n and $l$ (i) $\mathrm{n}=4, l=1 ;$ (ii) $\mathrm{n}=4, l=0$; (iii) $\mathrm{n}=3$, $l=2 ;$ (iv) $\mathrm{n}=3, l=1$ can be placed in order of increasing energy from the lowest to highest as
1 (iv) $<$ (ii) $<$ (iii) $<$ (i)
2 (ii) $<$ (iv) $<$ (i) $<$ (iii)
3 (i) $<$ (iii) $<$ (ii) $<$ (iv)
4 (iii) $<$ (i) $<$ (iv) $<$ (ii)
Explanation:
The greater is the value of $(\mathrm{n}+1)$, the greater si the energy of orbitals. (i) $\mathrm{n}=4, \mathrm{l}=1=4 \mathrm{p}$ orbital (ii) $\mathrm{n}=4, \mathrm{l}=0=4$ s orbital (iii) $\mathrm{n}=3, \mathrm{l}=2=3 \mathrm{~d}$ orbital (iv) $\mathrm{n}=3, \mathrm{l}=1=3 \mathrm{p}$ orbital According to the Aufbau principle, energies of abovementioned orbitals are in the order of - The increasing order of energy (iv) $3 \mathrm{p}<$ (ii) $4 \mathrm{~s}<$ (iii) $3 \mathrm{~s}<$ (i) $4 \mathrm{p}$
238924
According to Bohr's theory, the angular momentum for an electron in $5^{\text {th }}$ orbit is
1 $2.5 \mathrm{~h} / \pi$
2 $5 \mathrm{~h} / \pi$
3 $25 \mathrm{~h} / \pi$
4 $5 \pi / 2 \mathrm{~h}$
Explanation:
The angular momentum $(\ell)$ of an electron in a Bohr's orbit is given as $\mathrm{L}=\frac{\mathrm{nh}}{2 \pi}$ It is an integral multiple of $\frac{\mathrm{h}}{2 \pi}$ In the fifth Bohr orbit, the angular momentum of electron is $(\mathrm{L})=\frac{5 \mathrm{~h}}{2 \pi}$ $\begin{aligned} & \mathrm{L}=\frac{2.5 \mathrm{~h}}{\pi} \\ & =2.5 \mathrm{~h} \backslash \pi \end{aligned}$
UPTU/UPSEE-2010
Structure of Atom
238911
Assertion: $d^5$ configuration is more stable than $\mathrm{d}^4$ Reason: $d^5$ has more exchange energy as compared to $\mathrm{d}^4$ because $10 \& 6$ exchanges are possible in $\mathrm{d}^5 \& \mathrm{~d}^4$ respectively.
1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.
3 If Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
: $d^5$ configuration has exactly haif-filled configuration so has more exchange energy. While $\mathrm{d}^4$ has only four such (unpaired) electron these has less exchange energy. Maximum exchange energy leads to the stabilization of the atoms. Therefore, $d^5$ configuration is more stable them that of $\mathrm{d}^4$ configuration.
AIIMS 26 May 2019 (Morning)
Structure of Atom
238918
If an element with atomic number $Z=115$ has been discovered today in which of the following family would it had been placed and identify its electronic configuration. #[Qdiff: Very Easy, QCat: Theory Based, examname: Shift-I , AP- EAPCET-07-09-2021]#
Atomic number $\mathrm{Z}=115$ is Moscovium (Mc) is a synthetic chemical element and an extremely radioactive element. It belongs to nitrogen family and electronic configuration is $115^{\mathrm{Mc}}=[\mathrm{Rn}] 5 \mathrm{f}^{14} 6 \mathrm{~d}^{10} 75^2 7 \mathrm{p}^3$
Structure of Atom
238926
The electrons indentified by quantum numbers n and $l$ (i) $\mathrm{n}=4, l=1 ;$ (ii) $\mathrm{n}=4, l=0$; (iii) $\mathrm{n}=3$, $l=2 ;$ (iv) $\mathrm{n}=3, l=1$ can be placed in order of increasing energy from the lowest to highest as
1 (iv) $<$ (ii) $<$ (iii) $<$ (i)
2 (ii) $<$ (iv) $<$ (i) $<$ (iii)
3 (i) $<$ (iii) $<$ (ii) $<$ (iv)
4 (iii) $<$ (i) $<$ (iv) $<$ (ii)
Explanation:
The greater is the value of $(\mathrm{n}+1)$, the greater si the energy of orbitals. (i) $\mathrm{n}=4, \mathrm{l}=1=4 \mathrm{p}$ orbital (ii) $\mathrm{n}=4, \mathrm{l}=0=4$ s orbital (iii) $\mathrm{n}=3, \mathrm{l}=2=3 \mathrm{~d}$ orbital (iv) $\mathrm{n}=3, \mathrm{l}=1=3 \mathrm{p}$ orbital According to the Aufbau principle, energies of abovementioned orbitals are in the order of - The increasing order of energy (iv) $3 \mathrm{p}<$ (ii) $4 \mathrm{~s}<$ (iii) $3 \mathrm{~s}<$ (i) $4 \mathrm{p}$
