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nekit [7.7K]
2 years ago
13

Identify the group of elements that corresponds to each of the following generalized electron configurations and indicate the nu

mber of unpaired electrons for each: (a) [Noble gas] ns^2 np^5 (b) [noble gas] ns^2 (n-1)d^2 (c) [noble gas] ns^2 (n-1)d^10 np^1 (d)[noble gas] ns^2 (n-2)f^6
Chemistry
1 answer:
Alja [10]2 years ago
3 0

Answer:

(a) [Noble gas] ns² np⁵: Number of unpaired electron is 1 and belongs to group 17 i.e. halogen group of the periodic table.

(b) [noble gas] ns² (n-1)d²: Number of unpaired electron is 2 and belongs to group 4 of the periodic table.

(c) [noble gas] ns² (n-1)d¹⁰ np¹: Number of unpaired electron is 1 and belongs to group 13 of the periodic table.

(d)[noble gas] ns² (n-2)f⁶ : Number of unpaired electron is 6 and belongs to group 8 of the periodic table.

Explanation:

In the Periodic table, the chemical elements are arranged in 7 rows, called periods and 18 columns, called groups. They are organized in increasing order of atomic numbers.

(a) [Noble gas] ns² np⁵

As the total number of electrons in the p-orbital is 5. Therefore, the number of unpaired electron is 1.

This element has 2 electrons in ns orbital and 5 electrons in np orbital. <u>So there are 7 valence electrons.</u>

Therefore, this element belongs to the group 17 i.e. halogen group of the periodic table.

(b) [noble gas] ns² (n-1)d²

As the total number of electrons in the d-orbital is 2. Therefore, the number of unpaired electrons is 2.

This element has 2 electrons in ns orbital and 2 electrons in (n-1)d orbital. So there are <u>4 valence electrons.</u>

Therefore, this element belongs to the group 4 of the periodic table.

(c) [noble gas] ns² (n-1)d¹⁰ np¹

As the total number of electrons in the p-orbital is 1. Therefore, the number of unpaired electron is 1.

This element has 2 electrons in ns orbital and 1 electron in np orbital. So there are <u>3 valence electrons</u>.

Therefore, this element belongs to the group 13 of the periodic table.

(d)[noble gas] ns² (n-2)f⁶

As the total number of electrons in the f-orbital is 6. Therefore, the number of unpaired electron is 6.

This element has 2 electrons in ns orbital and 6 electrons in (n-2)f orbital. So there are <u>8 valence electrons.</u>

Therefore, this element belongs to the group 8 of the periodic table.

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a helium balloon has a volume of 2.95 liters at 25 c. the volume of the ballon decreased to 2.25 l after it is placed outside on
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Answer:

The outside temperature is -45.8°C

Explanation:

When a gas keeps on constant its moles and its pressure, we can assume that volume will be increased or decreased as the T° (absolute T° in K).

V1 / T1 = V2 / T2

2.95L/298K = 2.25L / T2

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3 years ago
which factors are needed to determine the amount of heat absorbed by an aluminum cube after it is warmed? check all that apply
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Answer:

Explanation:

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4 0
3 years ago
1(a). What is the solubility of calcium chloride (CaCl2) at 5°C?
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Answer:

60 g/100 g water  

Explanation:

Find 5 °C on the horizontal axis.

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7 0
3 years ago
A 1.00 g sample of a metal X (that is known to form X ions in solution) was added to 127.9 mL of 0.5000 M sulfuric acid. After a
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<u>Answer:</u> The metal having molar mass equal to 26.95 g/mol is Aluminium

<u>Explanation:</u>

  • To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}     .....(1)

Molarity of NaOH solution = 0.5000 M

Volume of solution = 0.03340 L

Putting values in equation 1, we get:

0.5000M=\frac{\text{Moles of NaOH}}{0.03340L}\\\\\text{Moles of NaOH}=(0.5000mol/L\times 0.03340L)=0.01670mol

  • The chemical equation for the reaction of NaOH and sulfuric acid follows:

2NaOH+H_2SO_4\rightarrow Na_2SO_4+H_2O

By Stoichiometry of the reaction:

2 moles of NaOH reacts with 1 mole of sulfuric acid

So, 0.01670 moles of NaOH will react with = \frac{1}{2}\times 0.01670=0.00835mol of sulfuric acid

Excess moles of sulfuric acid = 0.00835 moles

  • Calculating the moles of sulfuric acid by using equation 1, we get:

Molarity of sulfuric acid solution = 0.5000 M

Volume of solution = 127.9 mL = 0.1279 L    (Conversion factor:  1 L = 1000 mL)

Putting values in equation 1, we get:

0.5000M=\frac{\text{Moles of }H_2SO_4}{0.1279L}\\\\\text{Moles of }H_2SO_4=(0.5000mol/L\times 0.1279L)=0.06395mol

Number of moles of sulfuric acid reacted = 0.06395 - 0.00835 = 0.0556 moles

  • The chemical equation for the reaction of metal (forming M^{3+} ion) and sulfuric acid follows:

2X+3H_2SO_4\rightarrow X_2(SO_4)_3+3H_2

By Stoichiometry of the reaction:

3 moles of sulfuric acid reacts with 2 moles of metal

So, 0.0556 moles of sulfuric acid will react with = \frac{2}{3}\times 0.0556=0.0371mol of metal

  • To calculate the molar mass of metal for given number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Mass of metal = 1.00 g

Moles of metal = 0.0371 moles

Putting values in above equation, we get:

0.0371mol=\frac{1.00g}{\text{Molar mass of metal}}\\\\\text{Molar mass of metal}=\frac{1.00g}{0.0371mol}=26.95g/mol

Hence, the metal having molar mass equal to 26.95 g/mol is Aluminium

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