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3 years ago

How did Ernest Rutherford change the atomic model?

Chemistry

1 answer:

nadya68 [22]3 years ago

6 0

Answer:

D. He showed that most of an atom's mass was located in the atom's

nucleus.

Explanation:

Ernest Rutherford changed the atomic model because of his experiment which was the gold foil experiment. A beam of alpha particles was aimed at a piece of gold foil, most particles passed through but some were scattered backward which showed that the middle of an atom (nucleus) is the where most of the mass is located.

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Identify the reagents represented by the letters in the above reaction scheme. (enter your answer as a string of letters without

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Write the charge and full ground-state electron configuration of the monatomic ion most likely to be formed by each:

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Answer :

(a) The charge and full ground-state electron configuration of the monatomic ion is, (+1) and $1s^22s^22p^63s^23p^64s^23d^{10}4p^6$

(b) The charge and full ground-state electron configuration of the monatomic ion is, (-3) and $1s^22s^22p^6$

(c) The charge and full ground-state electron configuration of the monatomic ion is, (-1) and $1s^22s^22p^63s^23p^64s^23d^{10}4p^6$

Explanation :

For the neutral atom, the number of protons and electrons are equal. But, they are unequal when the atoms present in the form of ions or the atom has some charges.

When an unequal number of electrons and protons then it leads to the formation of ionic species.

Ion : An ion is formed when an atom looses or gains electron.

When an atom looses electrons, it will form a positive ion known as cation.

When an atom gains electrons, it will form a negative ion known as anion.

(a) The given element is, Rb (Rubidium)

As we know that the rubidium element belongs to group 1 and the atomic number is, 37

The ground-state electron configuration of Rb is:

$1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^1$

This element will easily loose 1 electron and form $Rb^+$ ion which attain stable noble gas electronic configuration.

The full ground-state electron configuration of Rb ion is:

$1s^22s^22p^63s^23p^64s^23d^{10}4p^6$

(b) The given element is, N (Nitrogen)

As we know that the nitrogen element belongs to group 15 and the atomic number is, 7

The ground-state electron configuration of N is:

$1s^22s^22p^3$

This element will easily gain 3 electrons and form $N^{3-}$ ion which attain stable noble gas electronic configuration.

The full ground-state electron configuration of N ion is:

$1s^22s^22p^6$

(c) The given element is, Br (Bromine)

As we know that the bromine element belongs to group 17 and the atomic number is, 35

The ground-state electron configuration of Rb is:

$1s^22s^22p^63s^23p^64s^23d^{10}4p^5$

This element will easily gain 1 electron and form $Br^-$ ion which attain stable noble gas electronic configuration.

The full ground-state electron configuration of Br ion is:

$1s^22s^22p^63s^23p^64s^23d^{10}4p^6$

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