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3 years ago

Use the periodic table to classify each of the elements below.

Chemistry

2 answers:

Alika [10]3 years ago

5 0

Answer:

Explanation:

Cadmium:(Cd)

Cadmium is transition metal present in group twelve. It is soft metal and properties are similar to the other group members like zinc and mercury. Its atomic number is forty eight and have two valance electrons.

Electronic configuration:

Cd₄₈ = [Kr] 4d¹⁰ 5s²

Vanadium: (V)

It is present in group five. It is malleable and ductile transition metal. Its atomic number is twenty three. Vanadium have five valance electrons.

Electronic configuration:

V₂₃ =[Ar] 3d³ 4s²

Xenon: Xe

Xenon is present in group eighteen. It is noble gas. Its outer most valance shell is complete that's why it is inert. its atomic number is fifty four. Xenon have eight valance electrons.

Electronic configuration:

Xe₅₄ = [Kr] 4d¹⁰ 5s² 5p⁶

Iodine: (I)

Iodine is present in group seventeen. Its outer most valance shell have seven electrons. Iodine is the member of halogen family. It gain one electron to complete the octet. its atomic number is fifty three.

Electronic configuration:

I₅₃ = [Kr] 4d¹⁰ 5s² 5p⁵

Potassium: (K)

Potassium is present in group one. it is alkali metal. Its atomic number is nineteen. Its valance shell has one electron. Potassium loses its one valance electron and gets stable electronic configuration.

Electronic configuration:

K₁₉ = [Ar] 4s¹

Strontium: Sr

Strontium is present in group two. it is alkaline earth metal. its atomic number is thirty eight and have two valance electrons.

Electronic configuration:

Sr₃₈ = [Kr] 5s²

dalvyx [7]3 years ago

5 0

Answer:

cadmium (Cd): transition metal

vanadium (V): transition metal

xenon (Xe): noble gas

iodine (I): halogen

potassium (K): alkali metal

strontium (Sr): alkaline earth metal

Explanation:

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9 months ago

Carbon dioxide and helium which has a higher van der waals force of attraction between molecules

raketka [301]

Answer:

Carbon dioxide

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3 years ago

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3 years ago

Which of the following is a possible set of quantum numbers for an electron (n,l,m0,ms) ?

Mashutka [201]

Answer is: (3, 2, 0, -1/2).

The principal quantum number (n) is one of four quantum numbers which are assigned to each electron in an atom to describe that electron's state.

For principal quantum number n=3:

1) azimuthal quantum number (l) can be l = 0...n-1:

l = 0, 1, 2.

The azimuthal quantum number determines its orbital angular momentum and describes the shape of the orbital.

2) magnetic quantum number (ml) can be ml = -l...+l.

ml = -2, -1, 0, +1, +2.

Magnetic quantum number specify orientation of electrons in magnetic field and number of electron states (orbitals) in subshells.

3) the spin quantum number (ms), is the spin of the electron.

ms = +1/2, -1/2.

(1, 1, 0, +1/2) is not correct because orbital quantum number cannot be l = 1 for n = 1.

(2, 1, 2, +1/2) is not correct because magnetic quantum number cannot be ml = 2 for orbital quantum number l = 1.

(3, -2, 1, -1/2) is not correct because orbital quantum number cannot be l = -2 for principal quantum number n = 3.

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3 years ago