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3 years ago

Which element has an outer(ending) electron configuration of s2p4?

Chemistry

1 answer:

goblinko [34]3 years ago

5 0

The question is in complete, the complete question is:

Which element has an outer electron configuration of s 2 p4 ?

(A) Ca (B) Cr (C) Ge (D) Se

Answer:

(D) Se

Explanation:

The outer electronic configuration of an atom reflects the group to which it belongs in the periodic table. Each group has its unique outermost electron configuration and number of electrons found in the outermost shell of the atoms of elements in the group.

s 2 p4 is the unique outermost electron configuration of group 16 elements. Group 16 elements have six electrons in their outermost shell. If we consider the options, only Se is a member of group 16. Hence it is only Se that has the outermost electron configuration s 2 p4 .

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86.14 mL of an acid solution was needed to neutralize 30.24 mL of a base solution of unknown concentrations. A second trial is r

Airida [17]

Answer:

The correct answer is option B.

Explanation:

As given ,that 30.24 mL of base was neutralize by 86.14 mL of acid which means that moles of base present in 30.24 mL are neutralized by moles of acid present in 86.14 mL.

After dilution of base from 30.24 mL to 50.0 mL .Since, the moles of base are same in the solution as that of the moles in solution before dilution. Moles of acid require to neutralize the base after dilution will same as a that of present moles of acid present in 86.14 mL.

5 0

3 years ago

Gold has a density of 1,200 lb./ft. What is the density of gold in g/em? For conversion factors use I lb. 453.6 g, and l inch-2.

Jlenok [28]

<u>Answer:</u> The density of gold in $g/cm^3$ is $19.22g/cm^3$

<u>Explanation:</u>

Density is defined as the ratio of mass of the object and volume of the object. Mathematically,

$\text{Density}=\frac{\text{Mass of the object}}{\text{Volume of the object}}$

We are given:

Density of gold = $1200lb/ft^3$

Using conversion factors:

1 lb = 453.6 g

1 feet = 12 inches

1 inch = 2.54 cm

Converting given quantity into $g/cm^3$ , we get:

$\Rightarrow (\frac{1200lb}{ft^3})\times (\frac{453.6g}{1lb})\times (\frac{1ft}{12inch})^3\times (\frac{1inch}{2.54cm})^3\\\\\Rightarrow 19.22g/cm^3$

Hence, the density of gold in $g/cm^3$ is $19.22g/cm^3$

6 0

4 years ago

Pcl5 exist but ncl5 does not.why?

jeyben [28]

N is as electronegative as Cl is. So Cl doesnt react with O, N

6 0

3 years ago

Characteristics that can only be observed when the object changes to a point where new matter is formed

Eddi Din [679]

Chemical properties

7 0

3 years ago

Which reactants would lead to a spontaneous reaction?

balandron [24]

Answer: Option (b) is the correct answer.

Explanation:

The elements which have excess or deficiency of electrons will react readily.

Atomic number of Mn is 25 and electronic configuration of $Mn^{2+}$ is [Ar] $4s^{0}3d^{5}$ . This configuration is stable.

Atomic number of Cr is 24 and electronic configuration of $Cr$ is [Ar] $4s^{1}3d^{5}$ . This configuration is not stable.

Atomic number of Fe is 26 and electronic configuration of $Fe$ is [Ar] $4s^{2}3d^{6}$ . This configuration is stable.

Atomic number of Cu is 29 and electronic configuration of $Cu^{2+}$ is [Ar] $4s^{0}3d^{9}$ . This configuration is not stable.

Atomic number of Al is 13 and electronic configuration of Al is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{1}$ . This configuration is not stable.

Atomic number of Ba is 56 and electronic configuration of $Ba^{2+}$ is [Kr] $4d^{10}5s^{2}5p^{6}$ . This configuration is stable.

Atomic number of Mg is 12 and electronic configuration of $Mg^{2+}$ is $1s^{2}2s^{2}2p^{6}$ . This configuration is stable.

Atomic number of Sn is 50 and electronic configuration of Sn is [Kr] $4d^{10}5s^{2}5p^{2}$ . This configuration is stable.

Thus, we can conclude that out of the given options, only Fe and $Cu^{2+}$ reactants would lead to a spontaneous reaction as they have incomplete sub-shells. Therefore, in order to gain stability they will readily react.

8 0

3 years ago