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2 years ago

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Nitrogen (atomic number 7) and phosphorus (atomic number 15) belong to group 15 of the Periodic Table. Write the electronic conf

iguration of these two elements. Which of these will be more electronegative? Why?

2 answers:

Zanzabum2 years ago

8 0

Explanation:

The atomic number of nitrogen is 7 , so its electronic configuration will be 2 , 5
The atomic number of phosphorus is 15 , so its electronic configuration is 2 , 8, 5

<u>Nitrogen is more electronegative than phosphorus. </u>

<u>Reason</u><u> </u><u>:</u><u>-</u>

<u>From top to bottom down a group, electronegativity decreases</u><u>.</u>

<u>Explanation</u><u> </u><u>:</u><u>-</u>

Number of shell increases down a group, so there is an increased distance b/w the valence electrons and nucleus, or a greater atomic radius.‎ ‎

Since <u>nitrogen</u> has a smaller atomic radius than <u>phosphorus</u>, the attraction of its nucleus towards the incoming electron is more than phosphorus. So , <u>because of smaller atomic size nitrogen accepts electrons more easily.</u>

scoundrel [369]2 years ago

3 0

Answer:

Electronic configuration of nitrogen: K=2 L=5

Electronic configuration of phosphorus : K = 2 L = 8 M = 5

The element nitrogen will be more electronegative because of the smaller size of its atom. nitrogen accepts electrons more easily.

Explanation:

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Answer:

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Explanation:

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2 years ago

1. The solubility of lead(II) chloride at some high temperature is 3.1 x 10-2 M. Find the Ksp of PbCl2 at this temperature.

solniwko [45]

Answer:

1) The solubility product of the lead(II) chloride is $1.2\times 10^{-4}$ .

2) The solubility of the aluminium hydroxide is $1.6\times 10^{-10} M$ .

3)The given statement is false.

Explanation:

1)

Solubility of lead chloride = $S=3.1\times 10^-2M$

$PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$

S 2S

The solubility product of the lead(II) chloride = $K_{sp}$

$K_{sp}=[Pb^{2+}][Cl^-]^2$

$K_{sp}=S\times (2S)^2=4S^3=4\times (3.1\times 10^{-2})^3=1.2\times 10^{-4}$

The solubility product of the lead(II) chloride is $1.2\times 10^{-4}$ .

2)

Concentration of aluminium nitrate = 0.000010 M

Concentration of aluminum ion = $1\timed 0.000010 M=0.000010 M$

Solubility of aluminium hydroxide in aluminum nitrate solution = $S$

$Al(OH)_3(aq)\rightleftharpoons Al^{3+}(aq)+3OH^-(aq)$

S 3S

The solubility product of the aluminium nitrate = $K_{sp}=1.0\times 10^{-33}$

$K_{sp}=[Al^{3+}][OH^-]^3$

$1.0\times 10^{-33}=(0.000010+S)\times (3S)^3$

$S=1.6\times 10^{-10} M$

The solubility of the aluminium hydroxide is $1.6\times 10^{-10} M$ .

3.

$Molarity=\frac{Moles}{Volume (L)}$

Mass of NaCl= 3.5 mg = 0.0035 g

1 mg = 0.001 g

Moles of NaCl = $\frac{0.0035 g}{58.5 g/mol}=6.0\times 10^{-5} mol$

Volume of the solution = 0.250 L

$[NaCl]=\frac{6.0\times 10^{-5} mol}{0.250 L}=0.00024 M$

1 mole of NaCl gives 1 mole of sodium ion and 1 mole of chloride ions.

$[Cl^-]=[NaCl]=0.00024 M$

Moles of lead (II) nitrate = n

Volume of the solution = 0.250 L

Molarity lead(II) nitrate = 0.12 M

$n=0.12 M]\times 0.250 L=0.030 mol$

1 mole of lead nitrate gives 1 mole of lead (II) ion and 2 moles of nitrate ions.

$[Pb^{2+}]=[Pb(NO_2)_3]=0.030 M$

$PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$

Solubility of lead(II) chloride = $K_{sp}=1.2\times 10^{-4}$

Ionic product of the lead chloride in solution :

$Q_i=[Pb^{2+}][Cl^-]^2=0.030 M\times (0.00024 M)^2=1.7\times 10^{-9}$

$Q_i$ ( no precipitation)

The given statement is false.

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3 years ago

How many electrons can occupy the 4p sublevel of a neutral atom of bromine

VMariaS [17]

Answer:

6

Explanation

every p sublevel holds up to 6 electrons

so the 4p sublevel must hold up to 6 electonsr

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2 years ago