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(a) True or false: The hydrogen atom is most stable when it has a full octet of electrons. (b) How many electrons must a sulfur atom gain to achieve an octet in its valence shell? (c) If an atom has the electron configuration \(1 s^{2} 2 s^{2} 2 p^{3},\) how many electrons must it gain to achieve an octet?

Short Answer

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(a) False, the hydrogen atom is stable with only 2 electrons in its 1s orbital. (b) A sulfur atom must gain 2 electrons to achieve an octet in its valence shell. (c) The atom with the given electron configuration, \(1s^{2}2s^{2}2p^{3}\), must gain 3 electrons to achieve an octet in its valence shell.

Step by step solution

01

Determine the stability of a hydrogen atom

The statement claims that the hydrogen atom is most stable when it has a full octet of electrons. However, the hydrogen atom needs only two electrons in its 1s orbital to achieve a stable electron configuration, as it has only one electron shell (the first shell) which can hold a maximum of just 2 electrons. Therefore, the statement is false. #b#
02

Determining the number of electrons a sulfur atom needs to gain for an octet

To find out how many electrons a sulfur atom must gain to achieve an octet in its valence shell, we must first understand its electron configuration. Sulfur is in Group 16 of the periodic table, which means it has six valence electrons (the electron configuration of sulfur is \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{4}\)). To achieve an octet, sulfur needs to have a total of eight valence electrons. By subtracting its current number of valence electrons (6) from the desired octet (8), we find out that sulfur must gain 2 electrons.
03

Answer for part (b)

So, a sulfur atom must gain 2 electrons to achieve an octet in its valence shell. #c#
04

Determining the number of electrons an atom needs to gain for an octet given an electron configuration

The given electron configuration for the atom is \(1s^{2}2s^{2}2p^{3}\). The valence shell is the last shell with electrons and in this case, it is the second shell (2s and 2p orbitals). The second shell can hold a maximum of 8 electrons.
05

Calculate missing electrons to achieve an octet

By summing the electrons present in the second shell (2 + 3 = 5) and comparing it against the desired octet (8), we can calculate the number of electrons needed to achieve an octet: 8 - 5 = 3.
06

Answer for part (c)

Therefore, the atom with the given electron configuration must gain 3 electrons to achieve an octet in its valence shell.

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Most popular questions from this chapter

The Ti \(\mathrm{Ti}^{2+}\) ion is isoelectronic with the Ca atom. (a) Write the electron configurations of \(\mathrm{Ti}^{2+}\) and (b) Calculate the number of unpaired electrons for Ca and for \(\mathrm{Ti}^{2+} .\) (c) What charge would Ti have to be isoelectronic with \(\mathrm{Ca}^{2}+?\)

There are many Lewis structures you could draw for sulfuric acid, \(\mathrm{H}_{2} \mathrm{SO}_{4}\) (each \(\mathrm{H}\) is bonded to an O). (a) What Lewis structure(s) would you draw to satisfy the octet rule? (b) What Lewis structure(s) would you draw to minimize formal charge?

Draw the Lewis structures for each of the following molecules or ions. Identify instances where the octet rule is not obeyed; state which atom in each compound does not follow the octet rule; and state how many electrons surround these atoms: (a) \(\mathrm{NO},(\mathbf{b}) \mathrm{BF}_{3},(\mathbf{c}) \mathrm{ICl}_{2}^{-},(\mathbf{d}) \mathrm{OPBr}_{3}(\) the \(\mathrm{P}\) is the central atom), (e) XeF.

Which of the following statements about electronegativity is false? (a) Electronegativity is the ability of an atom in a molecule to attract electron density toward itself. (b) Electronegativity is the same thing as electron affinity. (c) The numerical values for electronegativity have no units. (d) Fluorine is the most electronegative element. (e) Cesium is the least electronegative element.

Ammonia reacts with boron trifluoride to form a stable compound, as we saw in Section 8.7 . (a) Draw the Lewis structure of the ammonia-boron trifluoride reaction product. (b) The B-N bond is obviously more polar than the \(\mathrm{C}-\mathrm{C}\) bond. Draw the charge distribution you expect on the \(\mathrm{B}-\mathrm{N}\) bond within the molecule (using the delta plus and delta minus symbols mentioned in Section 8.4\()\) . ( ) Boron trichloride also reacts with ammonia in a similar way to the trifluoride. Predict whether the \(B-N\) bond in the trichloride reaction product would be more or less polar than the \(B-N\) bond in the trifluoride product, and justify your reasoning.

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