Give the formula of a negative ion that would have the same number of electrons as each of the following positive ions. a. Na + b. Ca 2+ c. Al 3+ d. Rb +
Give the formula of a negative ion that would have the same number of electrons as each of the following positive ions.
a. Na+
b. Ca2+
c. Al3+
d. Rb+
(a)
Expert Solution
Interpretation Introduction
Interpretation: The formula of negative ions that have the same number of electrons as each of the given positive ions is to be written.
Concept introduction: The ionic compound can be separated into ions. Ions are of two types. Positively charged ions are called cation and negatively charged ions are called anions.
To determine: The formula of negative ions that have the same number of electrons as in
Na+.
Answer to Problem 53E
The formula of negative ions is
F−,O2− and
N3−.
Explanation of Solution
The electronic configuration of
Na+ is,
1s22s22p6
It contains 10 electrons.
The atomic number of oxygen
(O) is 8 which means it has 8 electrons and its electronic configuration is,
1s22s22p4
It gains two electrons to get 10 electrons and form
O2− and its electronic configuration becomes,
1s22s22p6
The atomic number of fluorine
(F) is 9 which means it has 9 electrons and its electronic configuration is,
1s22s22p5
It gains an electron to get 10 electrons and form
F− and its electronic configuration becomes,
1s22s22p6
The atomic number of nitrogen
(N) is 7 which means it has 7 electrons and its electronic configuration is,
1s22s22p5
It gains three electrons to get 10 electrons and form
N3− and its electronic configuration becomes,
1s22s22p6
(b)
Expert Solution
Interpretation Introduction
Interpretation: The formula of negative ions that have the same number of electrons as each of the given positive ions is to be written.
Concept introduction: The ionic compound can be separated into ions. Ions are of two types. Positively charged ions are called cation and negatively charged ions are called anions.
To determine: The formula of negative ions that have the same number of electrons as in
Ca2+.
Answer to Problem 53E
The formula of negative ions is
Cl−,S2−andP3−.
Explanation of Solution
The electronic configuration of
Ca2+ is,
1s22s22p63s23p6
It contains 18 electrons.
The atomic number of chlorine
(Cl) is 17 which means it has 17 electrons and its electronic configuration is,
1s22s22p63s23p5
It gains an electron to get 18 electrons and form
Cl− and its electronic configuration becomes,
1s22s22p63s23p6
The atomic number of sulfur
(S) is 16 which means it has 16 electrons and its electronic configuration is,
1s22s22p63s23p4
It gains two electrons to get 18 electrons and form
S2− and its electronic configuration becomes,
1s22s22p63s23p6
The atomic number of phosphorous
(P) is 15 which means it has 15 electrons and its electronic configuration is,
1s22s22p63s23p3
It gains three electrons to get 18 electrons and form
P3− and its electronic configuration becomes,
1s22s22p63s23p6
(c)
Expert Solution
Interpretation Introduction
Interpretation: The formula of negative ions that have the same number of electrons as each of the given positive ions is to be written.
Concept introduction: The ionic compound can be separated into ions. Ions are of two types. Positively charged ions are called cation and negatively charged ions are called anions.
To determine: The formula of negative ions that have the same number of electrons as in
Al3+.
Answer to Problem 53E
The formula of negative ions is
F−,O2−andN3−.
Explanation of Solution
The electronic configuration of
Al3+ is,
1s22s22p63s23p1
It contains 10 electrons.
The atomic number of fluorine
(F) is 9 which means it has 9 electrons and its electronic configuration is,
1s22s22p5
It gains an electron to get 10 electrons and form
F− and its electronic configuration becomes,
1s22s22p6
The atomic number of oxygen
(O) is 8 which means it has 8 electrons and its electronic configuration is,
1s22s22p4
It gains two electrons to get 10 electrons and form
O2− and its electronic configuration becomes,
1s22s22p6
The atomic number of nitrogen
(N) is 7 which means it has 7 electrons and its electronic configuration is,
1s22s22p5
It gains three electrons to get 10 electrons and form
N3− and its electronic configuration becomes,
1s22s22p6
(d)
Expert Solution
Interpretation Introduction
Interpretation: The formula of negative ions that have the same number of electrons as each of the given positive ions is to be written.
Concept introduction: The ionic compound can be separated into ions. Ions are of two types. Positively charged ions are called cation and negatively charged ions are called anions.
To determine: The formula of negative ions that have the same number of electrons as in
Rb+.
Answer to Problem 53E
The formula of negative ions is
Br−,Se2−andAs3−.
Explanation of Solution
The electronic configuration of
Rb+ is,
1s22s22p63s23p63d104s24p6
It contains 36 electrons.
The atomic number of bromine
(Br) is 35 which means it has 35 electrons and its electronic configuration is,
1s22s22p63s23p63d104s24p5
It gains an electron to get 36 electrons and form
Br− and its electronic configuration becomes,
1s22s22p63s23p63d104s24p6
The atomic number of selenium
(Se) is 34 which means it has 34 electrons and its electronic configuration is,
1s22s22p63s23p63d104s24p4
It gains two electrons to get 36 electrons and form
Se2− and its electronic configuration becomes,
1s22s22p63s23p63d104s24p6
The atomic number of arsenic
(As) is 33 which means it has 33 electrons and its electronic configuration is,
1s22s22p63s23p63d104s24p3
It gains three electrons to get 36 electrons and form
As3− and its electronic configuration becomes,
1s22s22p63s23p63d104s24p6
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Consider an ionic compound, MX3, composed of generic metal M and generic gaseous halogen X.
The enthalpy of formation of MX3 is Δ?∘f=−925 kJ/mol.
The enthalpy of sublimation of M is Δ?sub=175 kJ/mol.
The first, second, and third ionization energies of M are IE1=579 kJ/mol, IE2=1677 kJ/mol, and IE3=2479 kJ/mol.
The electron affinity of X is Δ?EA=−369 kJ/mol. (Refer to the hint).
The bond energy of X2 is BE=179 kJ/mol.
Determine the lattice energy of MX3.
Consider an ionic compound, MX2MX2, composed of generic metal MM and generic, gaseous halogen XX.
The enthalpy of formation of MX2MX2 is Δ?∘f=−923ΔHf∘=−923 kJ/mol.
The enthalpy of sublimation of MM is Δ?sub=131ΔHsub=131 kJ/mol.
The first and second ionization energies of MM are IE1=755IE1=755 kJ/mol and IE2=1364IE2=1364 kJ/mol.
The electron affinity of XX is Δ?EA=−329ΔHEA=−329 kJ/mol. (Refer to the hint).
The bond energy of X2X2 is BE=153BE=153 kJ/mol.
Determine the lattice energy of MX2MX2.
Δ?lattice=ΔHlattice=
kJ/mol
Suppose a chemist discovers a new metallic element and names it "Xtrinsium" (Xt).
Xt exhibits chemical behaviour similar to an alkaline earth.
Xt(s) + F2(g) → XtF2(s)
Lattice energy for XtF2
-2360. kJ/mol
First Ionization energy of Xt
520. kJ/mol
Second Ionization energy of Xt
936 kJ/mol
Electron affinity of F
-327.8 kJ/mol
Bond energy of F2
154 kJ/mol
Enthalpy of sublimation (atomization) of Xt
180. kJ/mol
Use the above data to calculate ΔH°f for Xtrinsium fluoride.
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