NEET Test Series from KOTA - 10 Papers In MS WORD
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Structure of Atom
238960
The total number of orbitals in the fifth energy level is #[Qdiff: Easy, QCat: Theory Based, examname: JIPMER-2009 , UP CPMT-2008]#
1 5
2 10
3 18
4 25
Explanation:
For fifth energy level The value of $n$ will be 5 Therefore, the total number of orbitals is given by $\mathrm{n}^2=$ $5^2=25$
Structure of Atom
238967
Which of the following are isoelectronic species? (i) $\mathbf{N H}_3$ (ii) $\mathrm{CH}_3^{+}$ (iii) $\mathrm{NH}_2^{-}$ (iv) $\mathrm{NH}_4^{+}$ Choose the correct answer from the codes given below.
1 (i), (ii), (iii)
2 (ii), (iii), (iv)
3 (i), (ii), (iv)
4 (i), (iii), (iv)
5 (ii), (iii)
Explanation:
Species Number of electrons (1) $\mathrm{NH}_3 \quad 7+1 \times 3=10$ (2) $\mathrm{CH}_3^{+} \quad 6+1 \times 3-1=8$ (3) $\mathrm{NH}_2^{-} \quad 7+1 \times 2+1=10$ (4) $\mathrm{NH}_4^{+} \quad 7+1 \times 4-1=10$ Since, (i) (iii) and (iv) have the equal number of electrons therefore these are called as isoelectronic species.
Kerala-CEE-2014
Structure of Atom
238973
4d, 5p, $5 f$ and $6 p$ orbital's are arranged in the order of decreasing energy. The correct option is
As we know, Greater the $(\mathrm{n}+1)$ values greater is the energy $(n+1)$ values for, $\begin{aligned} & 4 d=4+2=6 \\ & 5 p=5+1=6 \\ & 5 f=5+3=8 \\ & 6 p=6+1=7 \end{aligned}$ Order of energy would be $5 \mathrm{f}>6 \mathrm{p}>5 \mathrm{p}>4 \mathrm{~d}$
NEET-2009
Structure of Atom
238977
The compound in which cation is isoelectronic with anion is:
1 $\mathrm{NaCl}$
2 $\mathrm{CsF}$
3 $\mathrm{NaI}$
4 $\mathrm{K}_2 \mathrm{~S}$
Explanation:
$\mathrm{NaCl}$ : No of $\mathrm{e}^{-}$in $\mathrm{Na}^{+}=11-1=10$ No of $\mathrm{e}^{-}$in $\mathrm{Cl}^{-}=17+1=18$ CsF No of $\mathrm{e}^{-}$in $\mathrm{Cs}^{+}=55-1=54$ No of $\mathrm{e}^{-}$in $\mathrm{F}^{-}=9+1=10$ NaI No of $\mathrm{e}^{-}$in $\mathrm{Na}^{+} \Rightarrow 11-1=10$ No of $\mathrm{e}^{-}$in $\mathrm{I}^{-} \Rightarrow 53+1=54$ $\mathrm{K}_2 \mathrm{~S}$ No of $\mathrm{e}^{-}$in $\mathrm{K}^{+}=19-1=18$ No of $\mathrm{e}^{-}$in $\mathrm{S}^{2-}=16+2=18$ So, $\mathrm{K}_2 \mathrm{~S}$ anion has isoelectronic with cation
238960
The total number of orbitals in the fifth energy level is #[Qdiff: Easy, QCat: Theory Based, examname: JIPMER-2009 , UP CPMT-2008]#
1 5
2 10
3 18
4 25
Explanation:
For fifth energy level The value of $n$ will be 5 Therefore, the total number of orbitals is given by $\mathrm{n}^2=$ $5^2=25$
Structure of Atom
238967
Which of the following are isoelectronic species? (i) $\mathbf{N H}_3$ (ii) $\mathrm{CH}_3^{+}$ (iii) $\mathrm{NH}_2^{-}$ (iv) $\mathrm{NH}_4^{+}$ Choose the correct answer from the codes given below.
1 (i), (ii), (iii)
2 (ii), (iii), (iv)
3 (i), (ii), (iv)
4 (i), (iii), (iv)
5 (ii), (iii)
Explanation:
Species Number of electrons (1) $\mathrm{NH}_3 \quad 7+1 \times 3=10$ (2) $\mathrm{CH}_3^{+} \quad 6+1 \times 3-1=8$ (3) $\mathrm{NH}_2^{-} \quad 7+1 \times 2+1=10$ (4) $\mathrm{NH}_4^{+} \quad 7+1 \times 4-1=10$ Since, (i) (iii) and (iv) have the equal number of electrons therefore these are called as isoelectronic species.
Kerala-CEE-2014
Structure of Atom
238973
4d, 5p, $5 f$ and $6 p$ orbital's are arranged in the order of decreasing energy. The correct option is
As we know, Greater the $(\mathrm{n}+1)$ values greater is the energy $(n+1)$ values for, $\begin{aligned} & 4 d=4+2=6 \\ & 5 p=5+1=6 \\ & 5 f=5+3=8 \\ & 6 p=6+1=7 \end{aligned}$ Order of energy would be $5 \mathrm{f}>6 \mathrm{p}>5 \mathrm{p}>4 \mathrm{~d}$
NEET-2009
Structure of Atom
238977
The compound in which cation is isoelectronic with anion is:
1 $\mathrm{NaCl}$
2 $\mathrm{CsF}$
3 $\mathrm{NaI}$
4 $\mathrm{K}_2 \mathrm{~S}$
Explanation:
$\mathrm{NaCl}$ : No of $\mathrm{e}^{-}$in $\mathrm{Na}^{+}=11-1=10$ No of $\mathrm{e}^{-}$in $\mathrm{Cl}^{-}=17+1=18$ CsF No of $\mathrm{e}^{-}$in $\mathrm{Cs}^{+}=55-1=54$ No of $\mathrm{e}^{-}$in $\mathrm{F}^{-}=9+1=10$ NaI No of $\mathrm{e}^{-}$in $\mathrm{Na}^{+} \Rightarrow 11-1=10$ No of $\mathrm{e}^{-}$in $\mathrm{I}^{-} \Rightarrow 53+1=54$ $\mathrm{K}_2 \mathrm{~S}$ No of $\mathrm{e}^{-}$in $\mathrm{K}^{+}=19-1=18$ No of $\mathrm{e}^{-}$in $\mathrm{S}^{2-}=16+2=18$ So, $\mathrm{K}_2 \mathrm{~S}$ anion has isoelectronic with cation
238960
The total number of orbitals in the fifth energy level is #[Qdiff: Easy, QCat: Theory Based, examname: JIPMER-2009 , UP CPMT-2008]#
1 5
2 10
3 18
4 25
Explanation:
For fifth energy level The value of $n$ will be 5 Therefore, the total number of orbitals is given by $\mathrm{n}^2=$ $5^2=25$
Structure of Atom
238967
Which of the following are isoelectronic species? (i) $\mathbf{N H}_3$ (ii) $\mathrm{CH}_3^{+}$ (iii) $\mathrm{NH}_2^{-}$ (iv) $\mathrm{NH}_4^{+}$ Choose the correct answer from the codes given below.
1 (i), (ii), (iii)
2 (ii), (iii), (iv)
3 (i), (ii), (iv)
4 (i), (iii), (iv)
5 (ii), (iii)
Explanation:
Species Number of electrons (1) $\mathrm{NH}_3 \quad 7+1 \times 3=10$ (2) $\mathrm{CH}_3^{+} \quad 6+1 \times 3-1=8$ (3) $\mathrm{NH}_2^{-} \quad 7+1 \times 2+1=10$ (4) $\mathrm{NH}_4^{+} \quad 7+1 \times 4-1=10$ Since, (i) (iii) and (iv) have the equal number of electrons therefore these are called as isoelectronic species.
Kerala-CEE-2014
Structure of Atom
238973
4d, 5p, $5 f$ and $6 p$ orbital's are arranged in the order of decreasing energy. The correct option is
As we know, Greater the $(\mathrm{n}+1)$ values greater is the energy $(n+1)$ values for, $\begin{aligned} & 4 d=4+2=6 \\ & 5 p=5+1=6 \\ & 5 f=5+3=8 \\ & 6 p=6+1=7 \end{aligned}$ Order of energy would be $5 \mathrm{f}>6 \mathrm{p}>5 \mathrm{p}>4 \mathrm{~d}$
NEET-2009
Structure of Atom
238977
The compound in which cation is isoelectronic with anion is:
1 $\mathrm{NaCl}$
2 $\mathrm{CsF}$
3 $\mathrm{NaI}$
4 $\mathrm{K}_2 \mathrm{~S}$
Explanation:
$\mathrm{NaCl}$ : No of $\mathrm{e}^{-}$in $\mathrm{Na}^{+}=11-1=10$ No of $\mathrm{e}^{-}$in $\mathrm{Cl}^{-}=17+1=18$ CsF No of $\mathrm{e}^{-}$in $\mathrm{Cs}^{+}=55-1=54$ No of $\mathrm{e}^{-}$in $\mathrm{F}^{-}=9+1=10$ NaI No of $\mathrm{e}^{-}$in $\mathrm{Na}^{+} \Rightarrow 11-1=10$ No of $\mathrm{e}^{-}$in $\mathrm{I}^{-} \Rightarrow 53+1=54$ $\mathrm{K}_2 \mathrm{~S}$ No of $\mathrm{e}^{-}$in $\mathrm{K}^{+}=19-1=18$ No of $\mathrm{e}^{-}$in $\mathrm{S}^{2-}=16+2=18$ So, $\mathrm{K}_2 \mathrm{~S}$ anion has isoelectronic with cation
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Structure of Atom
238960
The total number of orbitals in the fifth energy level is #[Qdiff: Easy, QCat: Theory Based, examname: JIPMER-2009 , UP CPMT-2008]#
1 5
2 10
3 18
4 25
Explanation:
For fifth energy level The value of $n$ will be 5 Therefore, the total number of orbitals is given by $\mathrm{n}^2=$ $5^2=25$
Structure of Atom
238967
Which of the following are isoelectronic species? (i) $\mathbf{N H}_3$ (ii) $\mathrm{CH}_3^{+}$ (iii) $\mathrm{NH}_2^{-}$ (iv) $\mathrm{NH}_4^{+}$ Choose the correct answer from the codes given below.
1 (i), (ii), (iii)
2 (ii), (iii), (iv)
3 (i), (ii), (iv)
4 (i), (iii), (iv)
5 (ii), (iii)
Explanation:
Species Number of electrons (1) $\mathrm{NH}_3 \quad 7+1 \times 3=10$ (2) $\mathrm{CH}_3^{+} \quad 6+1 \times 3-1=8$ (3) $\mathrm{NH}_2^{-} \quad 7+1 \times 2+1=10$ (4) $\mathrm{NH}_4^{+} \quad 7+1 \times 4-1=10$ Since, (i) (iii) and (iv) have the equal number of electrons therefore these are called as isoelectronic species.
Kerala-CEE-2014
Structure of Atom
238973
4d, 5p, $5 f$ and $6 p$ orbital's are arranged in the order of decreasing energy. The correct option is
As we know, Greater the $(\mathrm{n}+1)$ values greater is the energy $(n+1)$ values for, $\begin{aligned} & 4 d=4+2=6 \\ & 5 p=5+1=6 \\ & 5 f=5+3=8 \\ & 6 p=6+1=7 \end{aligned}$ Order of energy would be $5 \mathrm{f}>6 \mathrm{p}>5 \mathrm{p}>4 \mathrm{~d}$
NEET-2009
Structure of Atom
238977
The compound in which cation is isoelectronic with anion is:
1 $\mathrm{NaCl}$
2 $\mathrm{CsF}$
3 $\mathrm{NaI}$
4 $\mathrm{K}_2 \mathrm{~S}$
Explanation:
$\mathrm{NaCl}$ : No of $\mathrm{e}^{-}$in $\mathrm{Na}^{+}=11-1=10$ No of $\mathrm{e}^{-}$in $\mathrm{Cl}^{-}=17+1=18$ CsF No of $\mathrm{e}^{-}$in $\mathrm{Cs}^{+}=55-1=54$ No of $\mathrm{e}^{-}$in $\mathrm{F}^{-}=9+1=10$ NaI No of $\mathrm{e}^{-}$in $\mathrm{Na}^{+} \Rightarrow 11-1=10$ No of $\mathrm{e}^{-}$in $\mathrm{I}^{-} \Rightarrow 53+1=54$ $\mathrm{K}_2 \mathrm{~S}$ No of $\mathrm{e}^{-}$in $\mathrm{K}^{+}=19-1=18$ No of $\mathrm{e}^{-}$in $\mathrm{S}^{2-}=16+2=18$ So, $\mathrm{K}_2 \mathrm{~S}$ anion has isoelectronic with cation