Problem 3
Question
The number of electrons, neutrons and protons in a species are equal to 10,8 and 8 respectively. The proper symbol of the species is (a) \({ }^{16} \mathrm{O}_{8}\) (b) \({ }^{18} \mathrm{O}_{8}\) (c) \({ }^{18} \mathrm{Ne}_{10}\) (d) \({ }^{16} \mathrm{O}_{8}^{6}=\)
Step-by-Step Solution
Verified Answer
The correct symbol is (a) \(^{16}\text{O}_{8}\).
1Step 1: Identify the Atomic Number
The atomic number is the number of protons in the nucleus of an atom. According to the problem, the species has 8 protons. Thus, the atomic number is 8. This identifies the element as oxygen (O).
2Step 2: Calculate the Mass Number
The mass number is the total number of protons and neutrons in an atom's nucleus. In this species, there are 8 protons and 8 neutrons. Thus, the mass number is \(8 + 8 = 16\).
3Step 3: Write the Isotopic Symbol
An isotopic symbol is written as \(^A_ZX\), where \(A\) is the mass number, \(Z\) is the atomic number, and \(X\) is the element symbol. For this species, the symbol is \({ }^{16} \text{O}_{8}\).
4Step 4: Ensure Charge Neutrality
The number of electrons matches the number of protons (10 electrons and 8 protons), indicating a net charge of \(+2\). Thus, the full isotopic symbol considering charge is \({ }^{16} \text{O}_{8}^{2+}\).
5Step 5: Compare with Given Options
Compare the isotopic symbol \({ }^{16} \text{O}_{8}^{2+}\) with the given options. Since none of the provided options show the charge, we look for \({ }^{16} \text{O}_{8}\), which is option (a).
Key Concepts
Atomic NumberMass NumberIsotopes
Atomic Number
The atomic number is a fundamental characteristic of an atom, representing the number of protons found in its nucleus. It serves as a unique identifier for each chemical element, meaning no two different elements can have the same atomic number. For example, in the exercise, the atom in question has 8 protons, hence an atomic number of 8. This number directly identifies the element as oxygen (O), because every oxygen atom has 8 protons.
The atomic number not only helps in identifying the element, but it also determines the placement of the element in the periodic table. Elements are ordered by increasing atomic numbers, which also corresponds to their introduction of new electron shells. In summary, understanding atomic numbers is crucial in chemistry since it tells us what element we are dealing with just from knowing the number of protons in its nucleus.
The atomic number not only helps in identifying the element, but it also determines the placement of the element in the periodic table. Elements are ordered by increasing atomic numbers, which also corresponds to their introduction of new electron shells. In summary, understanding atomic numbers is crucial in chemistry since it tells us what element we are dealing with just from knowing the number of protons in its nucleus.
Mass Number
The mass number gives us a sum of the protons and neutrons within the nucleus of an atom. This number helps chemists in determining specific isotopes of an element, as opposed to the atomic number which only identifies the element type. The exercise mentions a species with 8 protons and 8 neutrons. Thus, its mass number is calculated by simply adding these two numbers together: \[ \text{Mass Number} = \text{Protons} + \text{Neutrons} = 8 + 8 = 16 \]This gives us a mass number of 16. Unlike the atomic number, the mass number can vary among atoms of the same element due to the presence of isotopes. So, while the mass number doesn't directly tell you which element it is, it indicates which isotope of that element you're looking at when combined with the atomic number.
Isotopes
Isotopes are atoms of the same element that have the same number of protons, hence the same atomic number, but differ in their number of neutrons and consequently their mass numbers. In simple terms, they are variants of the same element. For the element oxygen depicted in the exercise with an atomic number of 8, there are multiple isotopes such as
- \(^ {16} \text{O}\)
- \(^ {17} \text{O}\)
- \(^ {18} \text{O}\)
Other exercises in this chapter
Problem 1
How many e \(^{-}\)are present in p-sub-orbits in \(\mathrm{Zn}\)-atom? (a) 2 (b) 6 (c) 12 (d) 18
View solution Problem 2
Wavelength of spectral line emitted is inversely proportional to (a) energy (b) velocity (c) radius (d) quantum number
View solution Problem 5
Which one of the following forms a colourless solution in aqueous medium? (a) \(\mathrm{V}^{3+}\) (b) \(\mathrm{Cr}^{3+}\) (c) \(\mathrm{Ti}^{3+}\) (d) \(\mathr
View solution Problem 6
The incorrect configuration is (a) \(\mathrm{K}=[\mathrm{Ar}] 4 \mathrm{~s}^{1}\) (b) \(\mathrm{Cr}=[\mathrm{Ar}] 3 \mathrm{~d}^{5} 4 \mathrm{~s}^{1}\) (c) \(\m
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